Answer: A and D
a. Many objects move in arcs of circles or complete circles at times, and polar coordinates allowthe motion of such objects to be comprehended more easily. Some of Newton's laws in certaincases, such as calculation ofg from first principles, are much easier to calculate using polar coordinates.
b. "These coordinates are just used to confuse students.
c. ""Physics teachers are helping math teachers by getting students to practice theirtrigonometry.
d. ""In some cases, such as addition of forces, where a force magnitude is specified, it is simpler todescribe the forces in polar coordinates and be able to convert to the xy representation.
Explanation:
(A). Any force possessing a fixed magnitude and direction, it's always important to describe the for exactly as it is and also be able to convert it from one coordinate representation to another.
(D). Anything which involves radial motion(this is a motion along a radius) or motion along an arc of a circle or ellipse, this kind of motion is best explained and easily understood when in polar coordinates.
Someone plzzz helpppppp with this last question
Answer:
I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL
C IS CORRECT
A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of the tooth on the prey, in N/cm², is
a) 0.0013 N/cm²
b) 128 N/cm²
c) 320 N/cm²
d) 640 N/cm²
Answer:
640N/cm^2Answer D is correct
Explanation:
[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]
hope this helps
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good luck! have a nice day!
To move a large crate across a rough floor, you push on it with a force at an angle of 15 degrees below the horizontal. Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.49.
Answer:
663N
Explanation:
We need to find the force that will overcome the frictional force.
The angle of the normal force is 15°.
The mass of the crate is 32 kg
The coefficient of static friction is 0.49
Frictional force is given in terms of Normal force as:
F = μNcosθ
where μ = coefficient of static friction
N = normal force
θ = angle of normal force
Frictional force is given as:
F = mg
=>mg = μNcosθ
=> N = mg/(μcosθ)
N = (32 * 9.8) / (0.49 * cos15)
N= 313.6 / 0.473
N = 663 N
The force needed to cause the box to move must be 663N or greater.
first law of equilibrium
Answer:
for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.
Hope I helped
Answer:
An object in static equilibrium has zero net force acting upon it.
The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as
F = F1+ F2+ F3+ F4+. . . = 0
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.
mQ: the mass of the quarterback
mB: the mass of the football
(vQx)i: the horizontal velocity of quarterback before throwing the ball
(vBx)i: the horizontal velocity of football before being thrown
(vQx)f: the horizontal velocity of quarterback after throwing the ball
(vBx)f: the horizontal velocity of football after being thrown
Answer:
vBxf = 0.08625m/s
Explanation:
This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.
[tex]p_f=p_i[/tex]
pf: final momentum
pi: initial momentum
The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.
Then, you have:
[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex] (1)
mQ: the mass of the quarterback = 80kg
mB: the mass of the football = 0.43kg
(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s
(vBx)i: the horizontal velocity of football before being thrown = 0m/s
(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?
(vBx)f: the horizontal velocity of football after being thrown = 15 m/s
You replace the values of the variables in the equation (1), and you solve for (vBx)f:
[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]
Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.
Hence, the speed of the quarterback after he throws the ball is 0.08625m/s
cellus
An object ends up at a position of
327 m after a displacement of -144 m.
What was its initial position?
(Unit = m)
Answer:
Its initial position was 471 m.
Explanation:
We have,
Final position of the object is 327 m
Displacement of the object is -144 m
It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,
[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]
So, its initial position was 471 m.
A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.
Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
55.3
Explanation:
The computation of the number of bright-dark-bright fringe shifts observed is shown below:
[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]
where
d = [tex]3.95 \times 10^{-2}m[/tex]
[tex]\lambda = 400 \times 10^{-9}m[/tex]
n = 1.00028
Now placing these values to the above formula
So, the number of bright-dark-bright fringe shifts observed is
[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]
= 55.3
We simply applied the above formula so that the number of bright dark bright fringe shifts could come
where would you expect to find vesicles of neurotransmitters
A. Synaptic gap
B. postsynaptic dendrites
C. Channels in the postsynaptic
D. Presynaptic terminal button
Answer:
D. Presynaptic terminal button
explanation:
Terminal Buttons are small knobs at the end of an axon that release chemicals called neurotransmitters. The terminal buttons form the Presynaptic Neuron
hope this helped!
An 89.2-kg person with a density 1025 kg/m3 stands on a scale while completely submerged in water. What does the scale read?
Answer:
89.11kg
Explanation:
Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.
Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}
The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.
The weight of the fluid=0.087m3× 1kg/me = 0.087kg
Now the weight of the fluid displaced is referred to as the upthrust.
Now the real weight - the apparent weight = the upthrust.
Hence the apparent weight = real weight - upthrust
Apparent weight = 89.2-0.087 = 89.11kg
In the Life Cycle of Stars diagram, what stage does letter J represent?
A.) white dwarf
B.) black dwarf
C.) black hole
D.) neutron star
Which letters in the Life Cycle of Stars diagram represent stars on the main sequence?
A.) F & I
B.) C & G
C.) A & E
D.) B & D
In the Life Cycle of Stars diagram, what stage does letter L represent?
A.) neutron star
B.) black hole
C.) white dwarf
D.) black dwarf
In the Life Cycle of Stars diagram, what stage does letter I represent?
A.) neutron star
B.) black dwarf
C.) black hole
D.) white dwarf
In the Life Cycle of Stars diagram, what does letter D represent?
A.) a high mass star
B.) a white dwarf
C.) a cool star
D.) a low mass star
In the Life Cycle of Stars diagram, what stage does letter C represent?
A.) nuclear fusion
B.) a supernova
C.) a planetary nebula
D.) protostar formation
Which letter in the Life Cycle of Stars diagram represents a star forming region of space?
A.) M
B.) H
C.) J
D.) G
Which letter in the Life Cycle of Stars diagram represents a planetary nebula?
Group of answer choices
A.) G
B.) H
C.) L
D.) M
ANSWER: num 1 is black hole
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At what distance from a 70.0 Watt speaker is the intensity 0.0195 W/m^2
(Treat the speaker as point of the source)
(Unit=meters)
PLEASE HELP ME!
Answer:
Distance = 16.9 m
Explanation:
We are given;
Power; P = 70 W
Intensity; I = 0.0195 W/m²
Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;
I = Power/Unit area = P/(4πr²)
where;
P is the sound power
r is the distance.
Thus;
Making r the subject, we have;
r² = P/4πI
r = √(P/4πI)
r = √(70/(4π*0.0195))
r = √285.6627
r = 16.9 m
Answer:
16.9 m
Explanation:
The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
(a) Derive an expression for the induced emf in the inductor as a function of time.
(b) At t = 0, is the current through the inductor increasing or decreasing?
(c) At t = 0, is the induced emf opposing or aiding the flow of the charge carriers? (Remember that the direction of a positive induced emf is the same as the current direction and the direction of a negative induced emf is opposite the current direction.)
(d) How are the answers to parts b and c consistent with the behavior of inductors discussed in the text?
Answer:
(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) neither increasing or decreasing
(c) opposite to the flow of charge carriers
Explanation:
The current through an inductor of inductance L is given by:
[tex]I(t)=I_{max}sin(\omega t)[/tex] (1)
(a) The induced emf is given by the following formula
[tex]emf_L=-L\frac{dI}{dt}[/tex] (2)
You derivative the expression (1) in the expression (2):
[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) At t=0 the current is zero
(c) At t = 0 the emf is:
[tex]emf_L=-\omega LI_{max}[/tex]
w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.
(d) read the text carefully
At t zero, the current through the inductor neither increasing nor decreasing because current is zero.
The current through an inductor of inductance L can be calculated by
[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1
(a) The induced emf can be calculated by
[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2
Derivative the equation (1) in the equation (2)
[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]
(b) At t=0 the current is zero,
(c) At t = 0 the emf is:
[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]
Therefore, at t zero, the current through the inductor neither increasing nor decreasing because current is zero.
To know more about inductance,
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A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.
Answer:
d. 1.00 m
Explanation:
In 1905, Einstein proposed special theory of relativity of light.
This theory had a number of consequences or results. One of them is called "Length Contraction".
According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.
But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.
In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:
d. 1.00 m
Write the first equation of motion. Under what condition(s) is this equation valid?
Explanation:
The first equation of motion in kinematics is given by :
[tex]v=u+at[/tex] .....(1)
u is initial speed
a is acceleration
v is final speed
t is time
Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy is stored by a 50 ampere-minute 12
volt battery is approximately = 36,000 J = 36 kJ
Explanation:
Power in electrical circuits is given as
Power = IV
But power generally is defined as energy expended per unit time
Power = (Energy/time)
Energy = Power × Time
Energy = IV × Time
Energy = (I.t × V)
I.t = 50 Ampere-minute = 50 × 60 = 3000 Ampere-seconds
V = 12 V
Energy = 3,000 × 12 = 36,000 J = 36 kJ
Hope this Helps!!!
A rifle fires a 2.05 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 8.01 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 4.46 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
Answer:
Spring Constant = 279.58 N/m
Explanation:
We are given;
Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg
Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m
Maximum height; h = 4.46 m
The formula for the energy in the spring is given by;
E = ½kx²
where:
k is the spring constant
x is the distance the spring is compressed.
Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;
½kx² = mgh
Plugging in the relevant values, we have;
½ * k * 0.0801² = 0.0205 * 9.81 * 4.46
0.003208005k = 0.8969
k = 0.8969/0.003208005
k = 279.58 N/m
Two cars start moving from the same point. One travels south at 60 miyh and the other travels west at 25 miyh. At what rate is the distance between the cars increasing two hours later?
Answer:
65 m/h
Explanation:
Let the distance of the car moving south be y.
Let the distance of the car moving west be x.
Let the distance between the two cars be a.
These three distances can be represented as a right angled triangle. So we can say:
[tex]a^2 = x^2 + y ^2[/tex]
Let us differentiate with respect to time, since the distances are changing with respect to time:
[tex]2a\frac{da}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \\\\=>a\frac{da}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}[/tex]__________(1)
da/dt = rate of change of distance between two cars
The speed of the car moving south (dy/dt) is 60 m/h and the speed of the car moving west (dx/dt) is 25 m/h.
Therefore:
dy/dt = 60 m/h and dx/dt = 25 m/h
After two hours, the distance of the two cars will be:
y = 2 * 60 = 120 miles
x = 2 * 25 = 50 miles
Therefore:
[tex]a^2 = 50^2 + 120^2\\\\a^2 = 2500 + 14400 = 16900\\\\a = \sqrt{16900}\\ \\a = 130 miles[/tex]
From (1):
130(da/dt) = 50(25) + 120(60)
130(da/dt) = 1250 + 7200 = 8450
da/dt = 8450/130 = 65 m/h
Therefore, after two hours, the distance between the two cars is changing at a rate of 65 m/h.
16)
Gamma
rays
X-rays
UV
Infrared
Micro-
waves
Radio
waves
Visible light
Light is an electromagnetic wave and it has a place on the electromagnetic spectrum based on it energy and
wavelength. How does light's energy compare to the energy of other forms of electromagnetic waves?
A)
Light is less energetic than X-rays.
B)
Light is more energetic than X-rays.
Light is the least energetic electromagnetic wave.
D)
Light is the most energetic electromagnetic wave.
Answer:
Light is less energetic than X-rays.
Explanation:
The electromagnetic spectrum refers to the range of wavelengths or frequencies over which electromagnetic radiation extends. It is the entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light. In the electromagnetic spectrum, the entire distribution of electromagnetic radiation is done according to their frequency or wavelength.
The energy of an electromagnetic wave depends on its frequency and wavelength. The shorter the wavelength, the greater the energy of the electromagnetic wave but the larger frequency, the greater the energy of the electromagnetic wave.
X-rays has a frequency of about 1×10^20 Hz compared to visible light of frequency of about 1×10^15 Hz. Hence X-rays, having a larger frequency, is more energetic than visible light.
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, how many more revolutions will it rotate through in the next 5.00 s?
Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m
Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]
Calculate the inlet and exit velocity of water jet
[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]
The conservation of mass equation of steady flow
[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]
[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]
[tex]A_e\ \texttt {is the exit area of the jet}[/tex]
since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal
The expression for thickness of the jet
[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]
(b)
[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]
[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]
[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
9. How do air masses move?
Answer:
Air masses move with the global pattern of winds. In most of the United States, air masses generally move from west to east. They may move along with the jet stream in more complex and changing patterns.
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber, and from the bottom hangs a 10 gram lead weight. The density of lead is 11.3 g/cm3. What fraction of the bobber's volume is submerged, as a percent of the total volume
Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.
The given parameters;
diameter of the bobber, d = 4 cmmass of the bobber, m = 3 gmass of the lead, m = 10 gdensity of the lead, ρ = 11.3 g/cm³The volume of the bobber is calculated as follows;
[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]
The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;
[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]
The volume of the lead is calculated as follows;
[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]
The buoyant force experienced by the lead due to the volume submerged is calculated as follows
[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]
The total buoyant force is calculated as;
[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]
The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;
[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]
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An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelength. Calculate the current and voltage output when the detector is used in the photo-conductive and photovoltaic modes respectively. The reverse saturation current (Is) is 9 nA.
Answer:
I = 4.189 mA V = 0.338 V
Explanation:
In order to do this, we need to apply the following expression:
I = Is[exp^(qV/kT) - 1] (1)
However, as the junction of the diode is illuminated, the above expression changes to:
I = Iopt + Is[exp^(qV/kT) - 1] (2)
Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:
I ≅ Iopt
Therefore:
I ≅ I₀Aλq / hc (3)
Where:
I₀A = Area of diode (radiation)
λ: wavelength
q: electron charge (1.6x10⁻¹⁹ C)
h: Planck constant (6.62x10⁻³⁴ m² kg/s)
c: speed of light (3x10⁸ m/s)
Replacing all these values, we can get the current:
I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)
I = 4.189x10⁻³ A or 4.189 mA
Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:
I = Is[exp^(qV/kT) - 1]
k: boltzman constant (1.38x10⁻²³ J/K)
4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]
4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]
465,444.44 + 1 = exp(38.65V)
ln(465,445.44) = 38.65V
13.0508 = 38.65V
V = 0.338 V
Photoelectric effect:
A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white light, lambda=410-750nm?
B. The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7, and 4.5eV, respectively. Which of these metals will not emit electrons when visible light shines on it?
Answer:
A. K = 0.546 eV
B. cooper and iron will not emit electrons
Explanation:
A. This is a problem about photoelectric effect. Then you have the following equation:
[tex]K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi[/tex] (1)
K: kinetic energy of the ejected electron
Ф: Work function of the metal = 2.48eV
h: Planck constant = 4.136*10^{-15} eV.s
λ: wavelength of light = 410nm - 750nm
c: speed of light = 3*10^8 m/s
As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :
[tex]K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV[/tex]
B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm
[tex]E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV[/tex]
You compare the energies E1 and E2 with the work functions of the metals and you can conclude:
sodium = 2.3eV < E1
cesium = 2.1 eV < E1
cooper = 4.7eV > E1 (this metal will not emit electrons)
iron = 4.5eV > E1 (this metal will not emit electrons)
Identify the five categories of stressors.
Answer:
The five kinds of stressors are:
Acute time-limited
Brief naturalistic
Stressful events sequences
Chronic
Distant
Explanation:
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If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the inductance per unit length does not exceed 50 nH per meter? Express your answer using two significant figures.
Answer:
Inner radius = 2 mm
Explanation:
In a coaxial cable, series inductance per unit length is given by the formula;
L' = (µ/(2π))•ln(R/r)
Where R is outer radius and r is inner radius.
We are given;
L' = 50 nH/m = 50 × 10^(-9) H/m
R = 2.6mm = 2.6 × 10^(-3) m
Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m
Plugging in the relevant values, we have;
50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)
Rearranging, we have;
(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)
0.25 = ln((2.6 × 10^(-3))/r)
So,
e^(0.25) = (2.6 × 10^(-3))/r)
1.284 = (2.6 × 10^(-3))/r)
Cross multiply to give;
r = (2.6 × 10^(-3))/1.284)
r = 0.002 m or 2 mm
HELPP MEE
Which image illustrates the desired interaction of a sound wave with
soundproofing material in a recording studio?
Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.
What is the basis of soundproofing?Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.
Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.
Thus, the correct option is C, as the D image is showing the absorption.
For more details regarding soundproofing, visit:
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Answer: C.D
Explanation:...
How many significant figures does 0.09164500561 have?
Answer:
10 Sig Figs
Explanation:
Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.
Which statement BEST explains the relationship between voltage, current, and power?
A. If voltage increases and everything else remains constant, then power will increase.
B. If voltage increases and everything else remains constant, then power will decrease.
C. If current decreases and everything else remains constant, then power will increase.
D. Voltage and power are inversely related.
A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly off in opposite directions, each with speed 0.827c relative to an inertial reference frame S. What is the total energy of particle A?
Answer:
E = 389 MeV
Explanation:
The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,
Total Energy of A = E = Rest Mass Energy + Relative Energy
Using Einstein's Equation: E = mc²
E = m₀c² + mc²
From Einstein's Special Theory of Relativity, we know that:
m = m₀/[√(1-v²/c²)]
Therefore,
E = m₀c² + m₀c²/[√(1-v²/c²)]
E = m₀c²[1 + 1/√(1-v²/c²)]
where,
m₀c² = rest mass energy = 140 MeV
v = relative speed = 0.827 c
Therefore,
E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]
E = (140 MeV)(2.78)
E = 389 MeV
