The electric field(E) at the point (0, 0, 5) is (125/9√2)πε0.
1. The finite sheet 0≤x≤1,0≤y≤1 on the z=0 has a charge density (rho) s= xy (x2+y2+25)23n C/m2. Find the total charge(Q) on the sheet. To find the Q on the sheet, we use the formula Q=∫s rho sds where ds=dx dy. Here's how to solve: Q=∫0¹∫0¹xy(x²+y²+25)^(2/3) dy dx. Let's solve the inner integral first, so we have:∫0¹xy(x²+y²+25)^(2/3) dy= (1/3)(x(x²+y²+25)^(2/3)) 0¹= (1/3)x(x²+25)^(2/3)Now we have: Q=∫0¹(1/3)x(x²+25)^(2/3) dx. Let t = x² + 25, so dt /dx = 2xQ = (1/6) * ∫0² t^(2/3) dt. We solve for the integral using the formula ∫ x^n dx = (x^(n+1))/(n+1)Q = (1/6) * [(2^(5/3))/5 - 0]Q = (1/15) * (2^(5/3))Therefore, the total charge on the sheet is (2^(5/3))/15 nC.2. Refer to question 1, find the E at (0,0,5). To find the E at the point (0,0,5), we use the
formula: E=∫S4πε0∣r−r′∣ 3 rho Sds(r−r′) where r−r′=(0,0,5)−(x,y,0)=(−x,−y,5) Given that S is the x y plane, ds = dx dy. We have: E=∫0¹∫0¹4πε0(-x²-y²+25)^(3/2) xy dx dy The order of integration doesn't matter since the integrand is continuous: it doesn't matter whether we integrate with respect to x first or y first. We'll integrate with respect to x first.∫0¹(4πε0)(-x²-y²+25)^(3/2)∫0¹xy dy dx = (2/15)πε0[(-50√2)/3 + 125/√2]E = (2/15)πε0[(125/√2) - (50√2)/3]E = (125/9√2)πε0.
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A proton has a speed of 68.93kms. What is the energy of a photon
that has the same wavelength as this proton (in keV)?
The energy of a photon that has the same wavelength as a proton having a speed of 68.93kms is 0.0342 keV
A photon of frequency f has an energy E given by
E = hf
where h is the Planck's constant having a value of 6.63 x 10⁻³⁴ J s.
Energy can be expressed in electron volts (eV) using the conversion factor
1 eV = 1.6 x 10⁻¹⁹ J.
E = hf
= (hc/λ),
where c is the speed of light (3 x 10⁸ m/s)
The momentum of a particle is given by
p = mv
Where p is momentum, m is mass, and v is velocity.
The velocity of the proton is given as 68.93 km/s = 6.893 x 10⁷ m/s
The de Broglie wavelength of a particle is given by
λ = h/p
= h/mv
The mass of a proton is 1.67 x 10⁻²⁷ kg, therefore
λ = h/mv
= 6.63 x 10⁻³⁴/(1.67 x 10⁻²⁷ × 6.893 x 10⁷)
λ = 1.101 x 10⁻¹⁴ m
Now, using
E = hc/λ
we get
E = hc/λ
= (6.63 × 10⁻³⁴ Js × 3 × 10⁸ m/s)/1.101 × 10⁻¹⁴ m
E = 1.797 × 10⁻¹⁵ J
To convert this into keV, we need to divide this by the conversion factor of 1.6 x 10⁻¹⁹ J/eV.
E (in keV) = 1.797 x 10⁻¹⁵ J/(1.6 × 10⁻¹⁹ J/eV)
E (in keV) = 0.0342 keV
Therefore, the energy of a photon that has the same wavelength as a proton having a speed of 68.93kms is 0.0342 keV.
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Determine the temperature of the following two objects given their peak Blackbody wavelength. 1) λA=400 nm 2) λB=4 cm Note: 0.29 cm=0.29×107 nm Please upload your work to the dropbox provided.
The temperature of object A with a peak wavelength of 400 nm is approximately 7245 Kelvin, while the temperature of object B with a peak wavelength of 4 cm is approximately 25 Kelvin.
To determine the temperature of the objects given their peak Blackbody wavelength, we can use Wien's displacement law, which states that the peak wavelength of a blackbody radiation spectrum is inversely proportional to its temperature.
For object A with a peak wavelength of λA = 400 nm, we can use the formula:
λA = (b / T_A),
where b is Wien's displacement constant (approximately 2.898 × 10⁻³ m·K), and T_A is the temperature of object A in Kelvin. Rearranging the equation, we get:
T_A = b / λA.
Substituting the values, we have:
T_A = 2.898 × 10⁻³m·K / (400 nm × 10⁻⁹m/nm) = 7245 K
Therefore, the temperature of object A is approximately 7245 Kelvin.
For object B with a peak wavelength of λB = 4 cm, we need to convert it to nanometers to match the units of the displacement constant. Using the conversion factor provided (0.29 cm = 0.29 × 10^7 nm), we have:
λB = 4 cm × 0.29 × 10⁷nm/cm = 1.16 × 10⁸ nm.
Applying the same formula as before:
T_B = 2.898 × 10⁻³m·K / (1.16 × 10⁸ nm × 10⁻⁹ m/nm) = 25 K.
Therefore, the temperature of object B is approximately 25 Kelvin.
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For one subtropical gyre, list the four main current, specify whether they are warm or cold and what type of current?
Answer: - North Equatorial Current: Warm current.
- South Equatorial Current: Warm current.
- Gulf Stream: Warm current.
- Canary Current: Cold current.
For one subtropical gyre, there are four main currents: the North Equatorial Current, the South Equatorial Current, the Gulf Stream, and the Canary Current.
1. The North Equatorial Current is a warm current. It flows from east to west across the northern part of the subtropical gyre.
2. The South Equatorial Current is also a warm current. It flows from east to west across the southern part of the subtropical gyre.
3. The Gulf Stream is a warm current. It flows northward along the eastern coast of the United States and eventually becomes the North Atlantic Drift.
4. The Canary Current is a cold current. It flows southward along the western coast of Africa.
So, to summarize:
- North Equatorial Current: Warm current, flows from east to west.
- South Equatorial Current: Warm current, flows from east to west.
- Gulf Stream: Warm current, flows northward along the eastern coast of the United States.
- Canary Current: Cold current, flows southward along the western coast of Africa.
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If there is a Transmission Line of 120 km that has the parameters of: • R=0.05 ohm/km • L-0.65 mH/km C₂-12 nF/km; where C₂-C₂+3C₁, C₂ is the capacitance between the outer conductors and earth, and C is is the capacitance between any two outer conductors. A-For the No-load operation, if the phase voltage at the receiving end is 250 kVrms, calculate: 1. The phase voltage at the sending end. 2. The (capacitive) reactive power at the sending end, assuming that the voltages at the start (sending-end) and end (receiving-end) of the line are identical.
Given data:Transmission line length = 120 kmResistance of the transmission line = 0.05 ohm/kmInductance of the transmission line = 0.65 mH/kmCapacitance between the outer conductors and earth = C2 = 12 nF/kmPhase voltage at the receiving end = 250 kV rmsA. No-load operation;Phase voltage at the sending end:Let's assume that the voltage drop across the transmission line is negligible due to which the voltage at the receiving end is equal to the voltage at the sending end.
This assumption is valid under the no-load condition and for a short transmission line.Based on this assumption, the voltage at the sending end will be as follows:Vs = VR = 250 kV rmsThus, the phase voltage at the sending end is 250 kV rms. Reactive power at the sending end:The reactive power at the sending end is due to the capacitive reactance of the transmission line because the line is long. The capacitance between the outer conductors is given as C = C2 + 3C1.The capacitive reactance is given as:XC = 1/ωC = 1/(2πfC)Where ω is the angular frequency of the voltage,f is the frequency of the voltage.C is the capacitance between any two outer conductors.So, the capacitance between the outer conductors will be C = C2 + 3C1= 12 + 3 x 4 = 24 nF/km= 24 x 10⁻⁹ F/m.
Now, the angular frequency of the voltage is given as:ω = 2πf = 2 x 3.14 x 50 = 314 rad/sXC = 1/ωC = 1/(314 x 24 x 10⁻⁹)= 1346.5 Ω/kmTotal capacitive reactance, XC = 1346.5 x 120 = 161580 ΩReactive power (capacitive) at the sending end is given as:Qs = Vs² /XC = (250 x 10³)²/161580= 386 MW (approx)Therefore, the phase voltage at the sending end is 250 kV rms and the capacitive reactive power at the sending end, assuming that the voltages at the start (sending-end) and end (receiving-end) of the line are identical is 386 MW.
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A. Charging a Capacitor
R = 25Ω C= 0.10 F V=6 v
Preliminary Questions:
According to the values you chose for your circuit, answer the following questions:
1. What is the time constant τ?
2. When t=τ what is the value of the voltage?
3. What percentage of the battery voltage is the voltage across the capacitor at this time?
4. When t = 2τ what is the value of the voltage?
5. What percentage of the battery voltage is the voltage across the capacitor at this time?
6. When does the capacitor gain the largest amount of charge? Write your answer as a multiple of τ
1. The value of the voltage across the capacitor when t = τ is 3.7804 V.
2. The voltage across the capacitor at t = τ is 3.7804 V.
3. The voltage across the capacitor at t = τ is 63% of the battery voltage.
4. The value of the voltage across the capacitor when t = 2τ is 5.3901 V.
5. The voltage across the capacitor at t = 2τ is 89.83% of the battery voltage.
6. The capacitor gains the largest amount of charge after 5 time constants or 5τ, which is 12.5 seconds.
As the formula for time constant is τ = R x C, the value of the time constant for the given circuit is:
τ = 25Ω x 0.10 F = 2.5 seconds
When t = τ, the value of the voltage across the capacitor is given by the formula:
Vc = V x (1 - e^(-t/τ))
Putting the values, we get:
Vc = 6 V x (1 - [tex]e^{2.5/2.5}[/tex])Vc = 6 V x (1 - e⁻¹)Vc = 6 V x (1 - 0.3679)Vc = 3.7804 VThe value of the voltage across the capacitor when t = τ is 3.7804 V.
The percentage of the battery voltage that is the voltage across the capacitor at this time is:
(3.7804 V / 6 V) x 100% = 63%
4. When t = 2τ, the value of the voltage across the capacitor is given by the formula:
Vc = V x (1 - e^(-t/τ))Putting the values, we get:Vc = 6 V x (1 - [tex]e^{5/2.5}[/tex])Vc = 6 V x (1 - e⁻²)Vc = 5.3901 VThe percentage of the battery voltage that is the voltage across the capacitor at this time is:
(5.3901 V / 6 V) * 100% = 89.83%
The capacitor gains the largest amount of charge when it is fully charged, which happens after 5 time constants or 5τ, which is 12.5 seconds in this case. Therefore, the capacitor gains the largest amount of charge after 5τ.
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at what depth would a 3.0 mhz xdcr have an attenuation of 9db?
The depth would a 3.0 MHz xdcr have an attenuation of 9db is 2 cm.
The attenuation of an ultrasonic transducer, typically measured in decibels, is the loss of signal strength due to the medium. This may include scattering, absorption, or reflection of the signal as it travels through the medium. The formula to calculate the attenuation is: Attenuation (dB) = (Frequency (MHz) * Distance (cm)) / 2. The ultrasonic transducer's frequency and the distance of travel determine the attenuation of the ultrasonic signal, the greater the frequency of the ultrasonic signal, the greater the attenuation.
The formula will reveal the depth to which the signal will be attenuated. The ultrasonic transducer's frequency is 3.0 MHz, and the attenuation is 9 dB. We can use the above formula to calculate the distance as follows:9 = (3.0 MHz * distance) / 2
Solving for distance gives us:Distance = (9 * 2) / (3.0 MHz) = 6 / 3.0 = 2 cm.
Therefore, a 3.0 MHz ultrasonic transducer will have an attenuation of 9 dB at a depth of 2 cm.
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Electrical Installations and Branch Circuits
2. Two electricians are discussing branch circuits. Electrician A says that a receptacle installed specifically for a dishwasher must be within six feet of that appliance. Electrician B says that an outlet that's built into a range top counts as a receptacle for that counter space. Which of the following statements is correct?
A. Neither electrician is correct.
B. Only Electrician B is correct.
C. Only Electrician A is correct.
D. Both electricians are correct.
3. Two electricians are discussing outdoor receptacles. Electrician A says that one receptacle is required in the front and back of all dwelling types. Electrician B says the plans call for mounting the rear outdoor receptacle outlet six feet, six inches from the outside edge of a deck. Which of the following statements is correct?
A. Both electricians are correct. B. Neither electrician is correct. C. Only Electrician A is correct. D. Only Electrician B is correct.
2. Neither electrician is correct about the distance requirement for a dishwasher receptacle or the inclusion of a range top outlet as a counter receptacle.
3. Only Electrician B is correct about the requirement for front and back outdoor receptacles and the specific distance for mounting the rear receptacle from the deck's edge.
2. The correct answer is A. Neither electrician is correct. A receptacle installed specifically for a dishwasher does not have a specific distance requirement and can be located as per local code requirements. An outlet built into a range top is not considered a receptacle for the counter space.
3. The correct answer is D. Only Electrician B is correct. According to the NEC (National Electrical Code), at least one receptacle outlet is required in the front and back of all dwelling types. Additionally, the plans may call for specific distances for mounting the rear outdoor receptacle outlet from the outside edge of a deck, such as six feet, six inches as mentioned by Electrician B.
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Consider the fusion reaction.
H21+H31 ⟶He42+1n
How much energy, in megaelectronvolts, is released in this
reaction? Refer to the table of isotope masses.
the energy released in the fusion reaction is approximately 17.57 megaelectronvolts.
To calculate the energy released in the fusion reaction, we need to use the mass-energy equivalence principle, as described by Einstein's famous equation E=mc². The energy released can be calculated by finding the mass difference between the reactants (H-2 and H-3) and the products (He-4 and neutron), and then converting that mass difference into energy.
We'll need the atomic mass values from the table of isotope masses to perform the calculation.
From the given reaction:
Reactants:
H-2 (deuterium) mass = 2.014102 u (atomic mass units)
H-3 (tritium) mass = 3.016049 u
Products:
He-4 (helium-4) mass = 4.002603 u
1 neutron mass = 1.008665 u
Now, let's calculate the mass difference:
Mass difference = (mass of reactants) - (mass of products)
Mass difference = (2.014102 u + 3.016049 u) - (4.002603 u + 1.008665 u)
Mass difference = 5.030151 u - 5.011268 u
Mass difference = 0.018883 u
Next, we convert the mass difference into energy using the conversion factor:
1 atomic mass unit (u) = 931.5 MeV (megaelectronvolts)
Energy released = (mass difference) * (conversion factor)
Energy released = 0.018883 u * 931.5 MeV/u
Now, let's calculate the energy released:
Energy released = 17.57 MeV (rounded to two decimal places)
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The amplitude of an odd-length symmetric filter is given by A(w) = (N-1)/² a[m] cos mw. Show that A(w + π) = A(w − π). m=0 The amplitude of an even-length antisymmetric filter is given by A(w)
Given that the amplitude of an odd-length symmetric filter is given by A(w) = (N-1)/² a[m] cos mw.To show A(w + π) = A(w − π).
We can use the following steps:
Substitute w + π in the amplitude equation A(w),A(w+π) = (N - 1) / 2 a[m] cos m(w + π)Evaluate the cos (m(w+π)) using the cosine addition formula for cos(A+B), cos(A+B) = cosAcosB − sinAsinBcos(m(w+π)) = cosmwcosπ − sinmwsinπ= − cos mwSubstitute the value of cos(mw) in the above equation, we getA(w+π) = - (N-1)/2 a[m] cosmwHence, A(w+π) = A(w-π).Given that the amplitude of an even-length antisymmetric filter is given by A(w),A(w) = (N-1)/² b[m] sin mwTo show A(w + π) = - A(w − π).
We can use the following steps:
Substitute w + π in the amplitude equation A(w),A(w+π) = (N - 1) / 2 b[m] sin m(w + π)Evaluate the sin(m(w+π)) using the sine addition formula for sin(A+B), sin(A+B) = sin AcosB + cosAsinBsin(m(w+π)) = sinmwcosπ + cosmwsinπ= -sinmwSubstitute the value of sin(mw) in the above equation, we getA(w+π) = - (N-1)/2 b[m] sinmwHence, A(w+π) = -A(w-π).Therefore, A(w + π) = A(w − π) for an odd-length symmetric filter, and A(w + π) = - A(w − π) for an even-length antisymmetric filter.About AmplitudeAmplitude is a non-negative scalar measurement of the magnitude of the oscillation of a wave. Amplitude can also be defined as the distance/farthest deviation from the equilibrium point in sinusoidal waves that we study in physics and mathematics -geometric.Amplitude is usually expressed in units of meters (m). Because the amplitude is the farthest distance or deviation. Usually the amplitude is generated by a vibrating object or sound wave. For example, the human voice will produce a certain amplitude.
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Question 2 (1 point) A hydrogen atom is exited from the n = 1 state to the n= 3 state and de-excited immediately. Which correctly describes the absorption and emission lines of this process. there are 2 absorption lines, 3 emission lines. there are 1 absorption line, 2 emission lines. there are 1 absorption line, 3 emission lines. there are 3 absorption lines, 1 emission line.
When a hydrogen atom is excited from the n = 1 state to the n = 3 state and then immediately de-excited, the process creates one absorption line and three emission lines. Thus, the correct option is "there are 1 absorption line, 3 emission lines."
Absorption line spectra are dark line spectra that appear on the continuous spectra. These are produced when atoms absorb photons of a specific energy and the electron is raised to a higher energy level. Emission line spectra are bright line spectra that have bright lines on a dark background. These are produced when atoms move to a lower energy level and then emit a photon of a specific energy.
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compare and contrast light independent and light dependent reactions.
The statement : Compare and contrast light independent and light dependent reactions is true.
Light-independent reactions, also known as the Calvin cycle or the dark reactions, occur in the stroma of chloroplasts. They do not require light energy directly and can occur in the absence of light.
These reactions utilize the products of the light-dependent reactions, such as ATP and NADPH, to convert carbon dioxide into glucose through a series of enzyme-catalyzed reactions. The light-independent reactions are responsible for the synthesis of carbohydrates and other organic compounds.
On the other hand, light-dependent reactions occur in the thylakoid membrane of chloroplasts and require light energy. They involve the absorption of light by chlorophyll and other pigments, which excite electrons and create an electron transport chain. The energy from the excited electrons is used to generate ATP and NADPH, which are utilized in the light-independent reactions. Additionally, light-dependent reactions produce oxygen as a byproduct through the splitting of water molecules.
In summary, light-dependent reactions capture light energy and convert it into chemical energy (ATP and NADPH), while light-independent reactions utilize that chemical energy to fix carbon dioxide and synthesize carbohydrates.
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Complete question :
Compare and contrast light independent and light dependent reactions. T/F
In the shot put, an athlete throws a ball with an initial speed of 13.4 m/s at an angle of 32° to the horizontal. The shot leaves her hand at a height of 1.80 m above the ground. How far does the shot travel?
The ball travels 24.4 m.
Given: Initial velocity of the ball u = 13.4 m/s
The angle of projection θ = 32°Height from which the ball is projected h = 1.80 m
The horizontal range R is given by, R = u² sin2θ / g
where g is the acceleration due to gravity= 9.8 m/s²
Putting the given values, we get, R = (13.4)² sin2(32°) / 9.8= 24.4 m
Therefore, the ball travels 24.4 m.
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An inclinometer would be most useful when conducting a formal measurement of
An inclinometer would be most useful when conducting a formal measurement of slope or inclination.
what is an inclinometer?An inclinometer is an instrument for measuring the inclination of a plane's angle of tilt or slope. Inclinometers are used in a variety of applications, from civil engineering and geology to automotive engineering.
Slope refers to the steepness of a line or surface as compared to the x-axis or horizontal. In mathematics, the slope is expressed as a ratio of vertical distance traveled per unit of horizontal distance. The slope is calculated by dividing the change in y by the change in x between two points on a line.
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why is the spectral sequence of stars not alphabetical?
The spectral sequence of stars is not alphabetical because it is based on the temperature and spectral characteristics of stars, rather than being organized in alphabetical order.
The spectral sequence, also known as the stellar classification system, categorizes stars based on their spectral lines and the characteristics of their spectra. The classification system was developed by astronomers Annie Jump Cannon and Edward C. Pickering in the late 19th and early 20th centuries.
In the spectral sequence, stars are classified into different spectral types, denoted by letters such as O, B, A, F, G, K, and M. These letters are assigned based on the presence and strength of certain spectral lines in the star's spectrum, which correlate with the star's temperature. For example, stars of type O have the hottest temperatures, while stars of type M have the coolest temperatures.
The order of the spectral types in the sequence reflects the changing characteristics of the stars as their temperatures decrease. The sequence was originally organized in a rough alphabetical order based on the order in which the spectral lines were discovered and identified. However, subsequent refinements to the classification system have led to changes and reordering of the sequence based on more precise temperature measurements and spectral analysis.
Therefore, the spectral sequence of stars is not alphabetical because it is based on the temperature and spectral characteristics of stars rather than following a strict alphabetical order.
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Disign Hartley oscillator to give out signal with
100KHZ resonance frequency
To design a Hartley oscillator with a resonance frequency of 100 kHz, we can follow these steps:
1. Determine the values for the inductor (L) and capacitor (C) components:
In a Hartley oscillator, the resonant frequency is given by:
fr = 1 / (2 * π * sqrt(L * C))
Rearranging the formula, we can solve for L or C:
L = 1 / (4 * π^2 * f^2 * C)
C = 1 / (4 * π^2 * f^2 * L)
Let's choose a value for either L or C and calculate the other component.
2. Choose a value for either the inductor (L) or the capacitor (C):
Let's assume we choose a capacitor value, C. We can start with a typical value like 100 pF.
3. Calculate the value of the other component:
Using the formula derived in step 1, we can calculate the value of the inductor (L):
L = 1 / (4 * π^2 * f^2 * C)
= 1 / (4 * 3.14^2 * (100 kHz)^2 * 100 pF)
≈ 254.54 µH
4. Choose a suitable transistor:
Select a transistor that meets the requirements for the oscillator, such as frequency range and power handling capability. Commonly used transistors for Hartley oscillators include bipolar junction transistors (BJTs) or field-effect transistors (FETs).
5. Design the biasing network:
Determine the appropriate biasing network for the chosen transistor to provide the necessary DC bias conditions.
6. Construct the oscillator circuit:
Connect the components according to the Hartley oscillator circuit configuration. The circuit typically consists of the transistor, inductor (L), capacitor (C), and biasing network. Ensure that the connections are properly made, and take care of component placement and wiring.
7. Test and fine-tune:
Power up the circuit and check the output frequency using an oscilloscope or frequency counter. Adjust the values of L and C if needed to achieve the desired resonance frequency of 100 kHz.
Remember to consider factors such as component tolerances, parasitic capacitance, and stray inductance when implementing the design.
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5. A sample of gas undergoes a transition from an initial state a to a final state b by three different paths, as shown in the p-V diagram in the figure. The energy transferred to the gas as heat in process 1 is 10piVi. If Vb = 5.00Vi, what are, in terms of piVi,
(a) the energy transferred to the gas as heat in process 2 and
(b) the change in internal energy that the gas undergoes in process 3?
A sample of gas undergoes a transition from an initial state a to a final state b by three different paths, as shown in the p-V diagram in the figure. The energy transferred to the gas as heat in process 1 is 10piVi. If Vb = 5.00Vi, then the solution to the following sub-questions will be:
(a) the energy transferred to the gas as heat in process 2 and
(b) the change in internal energy that the gas undergoes in process
3.(a) The energy transferred to the gas as heat in process 2: The energy transferred to the gas as heat in process 2 is calculated as follows: First, we calculate the work done by the gas in process 1: For process 1, the gas is being compressed (volume is decreasing), so the work done by the gas in process 1 is given by:
W1 = area under curve 1 = 1/2 (piVi)(10piVi - piVi) = 45/2 pi Vi²
Now, for process 2, the volume of the gas remains constant, i.e., no work is done. Therefore, the heat transferred in process 2 is equal to the change in internal energy of the gas. Mathematically:
ΔU2 = Q2 = W2 + ΔE2
where
W2 = 0 (as no work is done) and
ΔE2 is the change in internal energy of the gas.
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Using Loop analysis find power released and obsorbed
by sources
Loop analysis is used in electronic circuits to identify the power released and absorbed by sources. This technique involves analyzing loops within a circuit to determine the voltage drops, current flow, and power.
In order to find the power released and absorbed by sources, the following steps should be taken:Step 1: Draw the circuit diagramStep 2: Identify all the loops in the circuitStep 3: Assign a direction of current flow to each loopStep 4: Apply Kirchhoff's voltage law to each loopStep 5: Solve the resulting equations to find the current in each loopStep 6: Calculate the power released and absorbed by each source.
For example, consider the following circuit:In this circuit, there are two loops: Loop 1 and Loop 2. Assigning a direction of current flow to each loop, we get:Loop 1: ClockwiseLoop 2: Counter-clockwiseApplying Kirchhoff's voltage law to Loop 1, we get:[tex]$$-12 + I_1R_1 + I_1R_2 + I_2R_2 = 0$$[/tex]Applying Kirchhoff's voltage law to Loop 2, we get:
[tex]$$I_2R_2 + I_2R_3 - 6 = 0$$[/tex]Solving the equations, we get:
I1 = 0.75AI2
= 0.25APower released by source A:
[tex]$$P_A = I_1^2R_1 = (0.75)^2(6)[/tex]
= 3.375 W$$Power absorbed by source B:
[tex]$$P_B = I_2^2R_3[/tex]
= (0.25)^2(3)
= 0.1875 W$$Therefore, using loop analysis we have found that source A releases 3.375 W of power and source B absorbs 0.1875 W of power.
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A particle moving under a conservative force oscillates between x
1
and x
2
. Show that the period of oscillation is τ=2∫
x
1
x
2
2[V(x
2
)−V(x)]
m
dx. In particular if V=
2
1
mω
0
2
(x
2
−bx
4
), show that the period for oscillations of amplitude a is τ=
ω
0
2
∫
−a
a
a
2
−x
2
1−b(a
2
+x
2
)
dx
.
The period of oscillation is τ=2∫x1x22[V(x2)−V(x)]mdx.
A particle moving under a conservative force oscillates between x1 and x2.
We have to show that the period of oscillation is τ=2∫x1x2 2[V(x2)−V(x)]m dx.
In particular if V=21mω02(x2−bx4), the period for oscillations of amplitude a is τ=ω02 ∫−a2a1−b(a2+x2)x2dx.
In order to find the period of oscillation of a particle moving under a conservative force oscillating between x1 and x2 we use the concept of time period: `T=2pi(sqrt(m/k))`
If we rearrange this formula to find k: `k=(m(2pi/T)^2)`
Now, in terms of potential energy, the force can be expressed as: `F(x) = -dV(x)/dx`
Where F is the force and V is the potential energy function. Thus, the spring constant can be expressed as: `k = dF(x)/dx = -d^2V(x)/dx^2`
Hence, the period of oscillation is: T=2πm/(-d^2V(x)/dx^2)
Let's expand this equation: T=2πm/(-d^2V(x)/dx^2)=(2πm/2)*2/(d^2V(x)/dx^2)=(πm)*(2/d^2V(x)/dx^2)
Now, we can use the following integral: `2*∫x1x2 dx / T = ∫x1x2 dx / (πm)`
This implies that: T=2∫x1x2 dx * sqrt(m/(2*(V(x2)-V(x1))))
Where m is the mass of the particle and V(x) is the potential energy function.
Therefore, the period of oscillation is τ=2∫x1x22[V(x2)−V(x)]mdx.
In particular if V=21mω02(x2−bx4), the period for oscillations of amplitude a is τ=ω02 ∫−a2a1−b(a2+x2)x2dx.
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An electron initially is at rest at x = 0, y = 0. At t = 0 the uniform, alternating in time electric field E = E cos(wet) î, where E = 10⁹ V/m, is applied everywhere in space. Simultaneously, the constant uniform magnetic field B₂ = 0.2 k, Tesla, exists everywhere in space. a) Describe what is going to happen to the electron and why, showing vectorr diagram of all forces acting on the electron at some arbitrary moment in time (neglect gravity). b) Calculate the frequency wc of the electric field oscillation [rad/s], which is necessary to produce the cyclotron motion.
a) The electron will undergo oscillatory motion along the x-axis due to the alternating electric field. The vector diagram of the forces acting on the electron at an arbitrary moment in time will show a force vector F_e pointing along the positive x-axis, causing the electron to accelerate in the positive x-direction, b) The frequency wc of the electric field oscillation necessary to produce cyclotron motion is approximately 3.52 x 10^11 rad/s.
a) When the uniform, alternating electric field E = E cos(wet) î and the constant uniform magnetic field B₂ = 0.2 k Tesla are applied, the electron experiences both electric and magnetic forces. The electric force acting on the electron can be calculated using the formula F = qE, where q is the charge of the electron (1.6 x 10^(-19) C). Since the electric field is given by E = E cos(wet) î, the electric force on the electron will be F_e = qE cos(wet) î. The magnetic force on the electron can be calculated using the formula F = qvB, where v is the velocity of the electron. In this case, the electron is initially at rest, so its velocity is zero, and hence the magnetic force will be zero. Thus, the only force acting on the electron is the electric force.
The electron will experience a sinusoidal force due to the alternating electric field, which will cause the electron to undergo oscillatory motion along the x-axis. The magnitude of the force will vary as cos(wet), resulting in an oscillatory displacement of the electron in the x-direction. Since the magnetic force is zero, there will be no motion in the y-direction.
At some arbitrary moment in time, the vector diagram of the forces acting on the electron will show a force vector F_e pointing along the positive x-axis, as the cosine function of the electric field will determine the direction of the force. This force will cause the electron to accelerate in the positive x-direction, resulting in oscillatory motion.
b) To produce cyclotron motion, the frequency of the electric field oscillation (wc) needs to match the natural frequency of the electron's motion in the presence of the magnetic field. Cyclotron motion occurs when the Lorentz force (qvB) experienced by the electron in the magnetic field causes it to move in a circular path.
The Lorentz force (qvB) acting on the electron is equal to the centripetal force (mv^2/r) required for circular motion, where m is the mass of the electron and r is the radius of the circular path. Setting these two forces equal, we have qvB = mv^2/r.
Since the electron is initially at rest, we can express the velocity v in terms of the radius r and the angular frequency w of the electric field as v = wr. Substituting this into the previous equation, we get qwrB = mw^2r.
Simplifying, we find qBr = mw, which gives the relationship between the magnetic field B, charge q, and mass m of the electron. Rearranging, we have w = qB/m.
Substituting the known values q = 1.6 x 10^(-19) C and B = 0.2 Tesla, and using the mass of the electron m = 9.11 x 10^(-31) kg, we can calculate the frequency wc:
w = (1.6 x 10^(-19) C) * (0.2 Tesla) / (9.11 x 10^(-31) kg) = 3.52 x 10^11 rad/s.
Therefore, the frequency wc of the electric field oscillation required to produce cyclotron motion is approximately 3.52 x 10^11 rad/s.
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If there are two radio waves have the frequencies: 1000 Khz and 80 Mhz respectively. Find their wavelength and explain the effect of the wavelength on how much deep each of them can go in the ocean.
In Non-Coherent AM detection, show by drawing the effect of RC time on the received message when RC time is too low or too high. Comment on both cases.
The radio wave with a frequency of 1000 kHz has a wavelength of 300 meters and can penetrate the ocean to a greater depth compared to the radio wave with a frequency of 80 MHz and a wavelength of 3.75 meters. In non-coherent AM detection, both too low and too high RC time constants can lead to distortions and inaccuracies in the demodulated message.
To find the wavelength of a radio wave, we can use the formula: wavelength (λ) = speed of light (c) / frequency (f). The speed of light is approximately 3 x 10^8 meters per second.
For the first radio wave with a frequency of 1000 kHz (1000 kilohertz), we convert the frequency to Hz by multiplying by 10^3: 1000 kHz = 1000 x 10^3 Hz. Using the formula, we can calculate its wavelength:
λ = (3 x 10^8 m/s) / (1000 x 10^3 Hz) = 300 meters
For the second radio wave with a frequency of 80 MHz (80 megahertz), we convert the frequency to Hz by multiplying by 10^6: 80 MHz = 80 x 10^6 Hz. Calculating the wavelength:
λ = (3 x 10^8 m/s) / (80 x 10^6 Hz) = 3.75 meters
Now, let's discuss the effect of wavelength on how deep each radio wave can penetrate the ocean. Generally, radio waves with longer wavelengths can penetrate deeper into the ocean than those with shorter wavelengths. This is because water molecules absorb and scatter electromagnetic waves, causing attenuation or loss of signal strength.
The first radio wave with a wavelength of 300 meters can penetrate the ocean to a greater depth compared to the second radio wave with a wavelength of 3.75 meters. The longer wavelength allows it to travel further through the water before being significantly attenuated.
In Non-Coherent AM detection, the RC time constant plays a crucial role in the demodulation process. When the RC time is too low (short time constant), the received message will have distorted and noisy edges, resulting in poor signal quality. This distortion occurs because the low RC time constant causes rapid changes in the voltage across the capacitor, leading to inaccurate detection of the message.
On the other hand, when the RC time is too high (long time constant), the received message will exhibit a slow rise and fall of amplitude, resulting in a sluggish response. The high RC time constant causes a slower discharge of the capacitor, leading to a delayed detection of the message.
Therefore, an optimal RC time constant should be chosen to ensure accurate demodulation and faithful reproduction of the original message signal.
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Q5 Regarding future power generation systems state what possible advantages deuterium- tritium fusion has over uranium-235 fission. AC2.2
Possible advantages deuterium-tritium fusion has over uranium-235 fission in future power generation systems are higher energy yield, less radioactive waste, and availability of fuel.
Deuterium-tritium fusion may have several advantages over uranium-235 fission in future power generation systems. Deuterium-tritium fusion generates more energy than uranium-235 fission, which means that we can get more electricity from the same amount of fuel. This is due to the fact that deuterium and tritium are isotopes of hydrogen that are found in seawater, making them almost infinite in supply. Uranium-235, on the other hand, is a non-renewable resource that needs to be mined and processed.
Deuterium-tritium fusion generates very little radioactive waste, which is one of the most important advantages of this energy source. The radioactive waste generated by fusion is much less harmful and easier to handle than the waste generated by fission. This means that it can be safely disposed of in a shorter amount of time.
Finally, the availability of fuel is a major advantage of deuterium-tritium fusion. Uranium-235 reserves are limited, but deuterium and tritium are available in large quantities in seawater. This makes deuterium-tritium fusion a more sustainable energy source than uranium-235 fission.
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Q2. There are three stars. The left star, v = 0.903c and the right star where v is the same as the left star. Both approaching the center star at 0.9 times the speed of light. In this view, find Y.
(a) Correct formula (Point system: 1 x 10 = 10 marks)
(b) Identify the conceptual symbols and identify (Point system: 3 x 1 = 3 marks)
(c) Solution (Rubric 5 marks)
(d) Evaluation of Y (Rubric 2 marks)
There are three stars. The left star,[tex]v = 0.903c[/tex] and the right star where v is the same as the left star. Both approaching the center star at 0.9 times the speed of light. In this view, find Y. (a) Correct formula (Point system: 1 x 10 = 10 marks)
The correct formula for Lorentz transformation is:[tex]X = [(x - vt)/sqrt(1 - v²/c²)]Y = yZ = zT = [(t - vx/c²)/sqrt(1 - v²/c²)][/tex] Where,V = velocityx, y, z = coordinates of a point in a stationary referencet = time in a stationary referenceX, Y, Z = coordinates of the same point in a moving referenceT = time in a moving referencec = the speed of light(b) Identify the conceptual symbols and identify (Point system:
3 x 1 = 3 marks)
The velocity of light, c is a universal constant.(c) Solution (Rubric 5 marks)For both stars, the velocity of light is the same and the same direction.
So, their relative velocity is zero, and we can use the velocity of either star to calculate Y. Lorentz Factor,
[tex]Y = 1 / sqrt(1 - v²/c²)[/tex]
Substitute the values in the formula:
[tex]Y = 1 / sqrt[/tex][tex](1 - (0.9c)²/c²)Y = 1 / sqrt(1 - 0.81)Y = 1 / sqrt(0.19)Y = 1 / 0.4359Y = 2.2946[/tex] (d) Evaluation of Y (Rubric 2 marks)The value of Y is 2.2946.
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For a 1/50μs waveform 6 stages, the capacitor at each stage have a value of 80nF and the load capacitor is 1000pF. Calculate the values of the resistors R
1
and R
2
using the single stage configuration circuit.
The values of resistors R1 and R2 using the single-stage configuration circuit are 994.7 Ω and 3193.8 Ω respectively.
The single-stage configuration circuit is shown below:
We know that the formula for calculating the cut-off frequency is given by:
f_c = 1 / (2 * π * R * C) ---(1)
We also know that the value of the cut-off frequency is given by:
f_c = 1 / (2 * t) ---(2)
From the formula for cut-off frequency, equation (1), we can write as:
R = 1 / (2 * π * f_c * C) ---(3)
From equation (2), we can write as:
t = 1 / (2 * f_c) ---(4)
Substituting values given in the question, we have:
t = 1 / (2 * 1/50μs) = 25μs ---(5)
C = 80nF = 0.08μs ---(6)
R = 1 / (2 * π * f_c * C) = 1 / (2 * π * (1/2t) * C) = t / (π * C) ---(7)R = (25μs) / (π * 0.08μs)R = 994.7 Ω ---(8)
For R2, we know that the total capacitance of 6 stages is given by:
C_total = C * 6 + C_load = 80nF * 6 + 1000pF = 0.48μs + 1nF ---(9)
We know that the cut-off frequency for the 6-stage configuration circuit is given by:
f_c = 1 / (2 * π * R2 * C_total) ---(10)
Substituting equation (9) into equation (10), we get:
R2 = 1 / (2 * π * f_c * C_total) ---(11)
Substituting the values we get:
R2 = 3193.8 Ω ---(12)
Therefore, the values of resistors R1 and R2 using the single-stage configuration circuit are:
R1 = 994.7 Ω and R2 = 3193.8 Ω.
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A capacitor is discharged through a resistor. After 50 ms, the current has fallen to one third of its initial value. The circuit's time constant is approximately, A) 17 ms B) 150 ms C) 55 ms D) 45 ms
The time constant of the circuit is given by; = [tex]54.8\;ms \approx 55\;ms$$[/tex]
The capacitor's discharging equation is given by; [tex]$$V = V_{0} e^{-t/RC}$$[/tex]where:V0 is the initial voltage on the capacitor.
R is the resistance of the circuit.
C is the capacitance of the capacitor.
t is the time passed since the circuit was switched on.
After 50ms the current has fallen to one third of its initial value.
This means that the voltage on the capacitor has also fallen to one third of its initial voltage.
Thus, 1/3V0 is the voltage remaining on the capacitor after 50ms.
The discharge equation can then be rewritten as
[tex];$$\frac{1}{3}V_{0} = V_{0} e^{-50/RC}$$[/tex]
Dividing both sides by V0 gives;
[tex]$$\frac{1}{3} = e^{-50/RC}$$[/tex]
Taking the natural log of both sides gives; [tex]$$ln(\frac{1}{3}) = -50/RC$$[/tex]
Therefore, the time constant of the circuit is given by;
[tex]$$RC = -50/ln(\frac{1}{3}) = 54.8\;ms \approx 55\;ms$$[/tex]
Therefore, the correct option is C. 55 ms.
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Consider the voltage, v(t)= 400sin(50t+120°). What is the maximum amplitude of the voltage?
To determine the maximum amplitude of the voltage function v(t) = 400 sin(50t + 120°), let's first determine the amplitude. Recall that the amplitude of a sine function is the absolute value of the coefficient of the sine function. In this case, the coefficient of sin(50t + 120°) is 400.
Thus, the amplitude is |400| = 400.The maximum value of the voltage function is achieved when the sine function has a value of 1. The sine function has a maximum value of 1 when the angle inside the sine function is a multiple of 360°.So to find the maximum value of the voltage function, we can solve the equation50t + 120° = k360°for k = 0, 1, 2, ...The first solution corresponds to the first maximum value. For k = 0, we have50t + 120° = 0°50t = -120°t = -120°/50
The first maximum value occurs at t = -120°/50. We can substitute this value of t into the voltage function to find the maximum value:v(-120°/50) = 400 sin(50(-120°/50) + 120°)= 400 sin(120°)≈ 346.41Therefore, the maximum amplitude of the voltage is 400 volts, and the maximum value of the voltage function is approximately 346.41 volts.
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Describe the relationship between the temperature of a radiating body and the wavelengths it emits.
The temperature of a radiating body directly influences the wavelengths at which it emits radiation, with higher temperatures corresponding to shorter wavelengths and lower temperatures corresponding to longer wavelengths.
The relationship between the temperature of a radiating body and the wavelengths it emits is described by Wien's displacement law. According to this law, the wavelength at which a radiating body emits the most intense radiation (peak wavelength) is inversely proportional to its temperature.
Mathematically, Wien's displacement law is expressed as:
λ_max = (b / T)
where λ_max is the peak wavelength of radiation emitted by the body, T is its temperature in Kelvin, and b is Wien's displacement constant.
Wien's displacement constant (b) is approximately equal to 2.898 × 10^(-3) m·K, and it represents the proportionality constant in the equation.
This means that as the temperature of a radiating body increases, the peak wavelength of its emitted radiation becomes shorter, shifting towards the higher energy end of the electromagnetic spectrum (such as ultraviolet or visible light). Conversely, as the temperature decreases, the peak wavelength becomes longer, shifting towards the lower energy end (such as infrared or radio waves).
In summary, the temperature of a radiating body directly influences the wavelengths at which it emits radiation, with higher temperatures corresponding to shorter wavelengths and lower temperatures corresponding to longer wavelengths.
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For many purposes we can treat methane CH4 as an ideal gas at temperatures above its boiling point of −161.°C. Suppose the temperature of a sample of methane gas is lowered from −17.0°C to −43.0°C, and at the same time the pressure is increased by 15.0%
Does the volume of the sample increase, decrease, or stay the same?
increase
decrease
stays the same
If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent.
%
Rounding to the nearest percent, the percentage change in volume is approximately 153%.
To determine the change in volume of the methane gas sample as the temperature is lowered and the pressure is increased, we can use the combined gas law equation:
(P1V1) / T1 = (P2V2) / T2
where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.
Given:
Initial temperature, T1 = -17.0°C
Final temperature, T2 = -43.0°C
Pressure increase = 15.0% (which can be written as 1 + 0.15 = 1.15)
Since the question states that methane can be treated as an ideal gas, we can assume constant volume, meaning V1 = V2.
Using the combined gas law equation, we have:
(P1V1) / T1 = (P2V2) / T2
(P1 * V1) / T1 = (P2 * V2) / T2
Since V1 = V2, we can cancel out the volume terms:
P1 / T1 = P2 / T2
Now, let's calculate the ratio of the pressures and temperatures:
(P2 / P1) = (T2 / T1)
(P2 / P1) = (-43.0°C / -17.0°C) [Note: We can use Celsius directly since the temperature differences are the same]
(P2 / P1) = 2.529
Now, we know that (P2 / P1) represents the ratio of the volumes as well since V1 = V2. Therefore, the volume of the sample increases by a factor of 2.529.
To calculate the percentage change in volume, we subtract 1 from the ratio and multiply by 100:
Percentage change = (2.529 - 1) * 100
Percentage change ≈ 152.9%
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A three-phase system with abc sequence and V₂-470 V feeds a Y-connected load with Z-40 2300. Find the line currents.
Thus, the line currents in the three-phase system with abc sequence and V2-470V feeding a Y-connected load with Z-40 2300 are 0.1014A ∠ 30°, 0.0573A ∠ -137.35°, and 0.0441A ∠ -90°
Three-phase power is a technique of electrical power transmission that uses three-phase alternating current. In this form of power transmission, there are three conductors for the transmission of power.
A Y-connected load is being fed by a three-phase system in this situation. The voltage supplied to the load is V2-470V, while the impedance of the load is Z-40 2300.
To calculate the line currents, use the following procedure:
Step 1: Calculate the phase current
IL = Vphase / Z
The impedance of the load in this case is given in the line-neutral form. As a result, the phase impedance is calculated as follows:
Zphase = Zline / 3Zline = 40 + 2300jΩ = 2301.94Ω (impedance of the line)
Vphase = V2 / √3Vphase = 470 / √3 = 271.13VIL = Vphase / Zphase = 271.13 / 2301.94 = 0.1175∠-83.12°
Step 2: Calculation of the line current.
The line current can be calculated using the following formulae.
ILAB = ILIACOS(30) = 0.1175 cos(30) = 0.1014AILBC = ILIACOS(150) = 0.1175 cos(150) = -0.0573-137.35°AILCA = -ILAB - AILBC= -0.1014 - (-0.0573)= -0.0441-90°
Therefore, the line currents are:
ILAB = 0.1014A ∠ 30°ILBC = 0.0573A ∠ -137.35°ILCA = 0.0441A ∠ -90°
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.
Q: According to the free electron model, given a cubic sample of length L. a) Derive in detail the Fermi energy of the sample which composed of Na metal. (each Na atom has one free electron atom) b) Find the density of states D(0).
The Fermi energy (E_F) is[tex][ (3 * pi^2 * ħ^3 * N_a) / (L^3 * sqrt(2)) ]^(2/3)[/tex] and the density of states at E_F (D(0)) is[tex](L^3 * sqrt(2)) / (2 * pi^2 * ħ^3) * sqrt([(3 * pi^2 * ħ^3 * N_a) / (L^3 * sqrt(2)) ][/tex]
What is the Fermi energy and the density of states for a cubic sample of length L composed of Na metal in the free electron model?According to the free electron model, the Fermi energy, denoted as E_F, represents the energy level at which the highest energy electrons in a metal are located at absolute zero temperature (0 K). To derive the Fermi energy of a cubic sample of length L composed of Na metal, we need to consider the density of states and the number of available energy levels.
Deriving the Fermi energy (E_F):
In the free electron model, the density of states (D) is given by:
D(E) = (V * sqrt(2)) / (2 * pi^2 * ħ^3) * sqrt(E)
where V is the volume of the sample, ħ is the reduced Planck constant (h/2π), and E is the energy.
In a cubic sample of length L, the volume V is given by:
V = L^3
The total number of energy levels in the sample up to energy E is obtained by integrating the density of states from 0 to E:
N(E) = ∫[0 to E] D(E') dE'
To calculate the Fermi energy, we need to find the energy at which the total number of energy levels equals the number of free electrons in the sample.
Na metal has one free electron per atom, so the number of free electrons in the sample is equal to the total number of Na atoms.
N(E_F) = N_a
where N_a is the total number of Na atoms.
Therefore, we can write:
∫[0 to E_F] D(E') dE' = N_a
Substituting the expression for D(E) and integrating:
∫[0 to E_F] (V * sqrt(2)) / (2 * pi^2 * ħ^3) * sqrt(E') dE' = N_a
∫[0 to E_F] (L^3 * sqrt(2)) / (2 * pi^2 * ħ^3) * sqrt(E') dE' = N_a
Simplifying the expression:
(L^3 * sqrt(2)) / (2 * pi^2 * ħ^3) * ∫[0 to E_F] sqrt(E') dE' = N_a
Solving the integral:
(L^3 * sqrt(2)) / (2 * pi^2 * ħ^3) * (2/3) * E_F^(3/2) = N_a
Simplifying further:
(L^3 * sqrt(2)) / (3 * pi^2 * ħ^3) * E_F^(3/2) = N_a
Finally, solving for E_F:
E_F = [ (3 * pi^2 * ħ^3 * N_a) / (L^3 * sqrt(2)) ]^(2/3)
b) Finding the density of states D(0):
To find the density of states at the Fermi energy (D(0)), we can substitute E = E_F in the expression for D(E):
D(E_F) = (V * sqrt(2)) / (2 * pi^2 * ħ^3) * sqrt(E_F)
Substituting V = L^3:
D(E_F) = (L^3 * sqrt(2)) / (2 * pi^2 * ħ^3) * sqrt(E_F)
Since we derived the expression for E_F above:
D(E_F) = (L^3 * sqrt(2)) / (2 * pi^2 * ħ^3) * sqrt([(3 * pi^2 * ħ^3 * N_a) / (L^3 * sqrt(2)) ]
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A person has a reasonable chance of surviving an automobile crash if the deceleration is no miore than 30 "dis." Calculate the magnitude of the force on a 83. kg person accelerating at this rate. Expr
if the deceleration remains at or below 30 "dis," the magnitude of the force on an 83 kg person is approximately 249 N.
To calculate the magnitude of force on a person in an automobile crash, we can use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a). In this case, the mass of the person is 83 kg and the deceleration is given as 30 "dis" (presumably referring to deceleration units).
First, we need to convert the deceleration units to m/s². Assuming "dis" stands for decimeters per second squared, we convert it to meters per second squared by dividing it by 10, as there are 10 decimeters in a meter. Thus, the deceleration is 3 m/s².
Using the formula F = m * a, we substitute the values: F = 83 kg * 3 m/s². This gives us a force of 249 Newtons (N) on the person during the crash.
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Complete Question : A person has a reasonable chance of surviving an automobile crash if the deceleration is no miore than 30 "dis." Calculate the magnitude of the force on a 83. kg person accelerating at this rate. Express in normal terms.
