Answer:
2500 N/m²
Explanation:
Pressure: This can be defined as the force acting normally on a surface per unit area.
The expression for pressure is give as
P = F/A...................... Equation 1
Where P = pressure (N/m²), F = force (N), A = Contact area (m²)
Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.
Substitute into equation 1
P = 500/0.2
P = 2500 N/m²
Hence the pressure exerted to the surface is 2500 N/m²
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.
Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet 2, and a laser beam is directed onto the sheets from above so that it strikes the interface at an angle of 26.50 with the normal. The refracted beam in sheet 2 makes an angle of 31.70 with the normal. The experiment is repeated with sheet 3 on top of sheet 2, and with the same angle of incidence, the refracted beam makes an angle of 36.70 with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, determine the expected angle of refraction in sheet 3? Assume the same angle of incidence.
Answer:
The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]\theta _i = 26.50 ^o[/tex]
The angle of refraction angle for sheet 1 is [tex]\theta _{r_1}} = 31.70 ^o[/tex]
The angle of refraction for sheet 3 is [tex]\theta _{r_3}} = 36.70 ^o[/tex]
According to Snell's law
[tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]
Where [tex]n_1 \ and \ n_2[/tex] are refractive index of sheet 1 and sheet 2
=> [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]
Also when sheet 3 in on top of sheet 2
[tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
substituting for [tex]n_2[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
=> [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]
when sheet 1 in on top of sheet 3
[tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3
substituting for [tex]n_3[/tex]
[tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
=> [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]
substituting values
[tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]
=> [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]
=> [tex]\theta_{r_s } = 23.1 ^o[/tex]
A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her
Answer:
The average upward force exerted by the water is 988.2 N
Explanation:
Given;
mass of the diver, m = 60 kg
height of the board above the water, h = 10 m
time when her feet touched the water, t = 2.10 s
The final velocity of the diver, when she is under the influence of acceleration of free fall.
V² = U² + 2gh
where;
V is the final velocity
U is the initial velocity = 0
g is acceleration due gravity
h is the height of fall
V² = U² + 2gh
V² = 0 + 2 x 9.8 x 10
V² = 196
V = √196
V = 14 m/s
Acceleration of the diver during 2.10 s before her feet touched the water.
14 m/s is her initial velocity at this sage,
her final velocity at this stage is zero (0)
V = U + at
0 = 14 + 2.1(a)
2.1a = -14
a = -14 / 2.1
a = -6.67 m/s²
The average upward force exerted by the water;
[tex]F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N[/tex]
Therefore, the average upward force exerted by the water is 988.2 N
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
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
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!
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.
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:
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
¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?
Answer:
La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h
Explanation:
La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.
Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?
[tex]distancia=\frac{1 segundo*90 km}{3600 segundos}[/tex]
distancia= 0.025 km
Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?
[tex]distancia=\frac{0.025 km*1,000 metros}{1 km}[/tex]
distancia= 25 metros
La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h
La cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.
Para obtener la velocidad de la motocicleta en un segundo, necesitaremos convertir 90 km / h en metros por segundo
Usando la tasa de conversión;
1000 m = 1 km
1 hora = 3600 segundos
[tex]\frac{90km}{hr} = \frac{90km \times 1000 m}{1km \times 3600s} \\\\\frac{90km}{hr} = \frac{90,000 m}{3600s} =25m/s \\[/tex]
Esto muestra que la cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.
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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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Identify the five categories of stressors.
Answer:
The five kinds of stressors are:
Acute time-limited
Brief naturalistic
Stressful events sequences
Chronic
Distant
Explanation:
yeah
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
Parallel light rays with a wavelength of 610nm fall on a single slit. On a screen 3.10m away, the distance between the first dark fringes on either side of the central maximum is 4.00mm.
What is the width of the slit?
Answer:
The width of the slit will be ".946 mm".
Explanation:
The given values are:
Wavelength = 610 × 10⁻⁹
Length, L = 3 m
As we know,
⇒ [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]
On applying cross-multiplication, we get
⇒ [tex]a=9.46\times 10^{-4}[/tex]
⇒ [tex]a = .946 mm[/tex]
Assume the three blocks (m. = 1.0 kg, m = 20 kg and m = 40 ko) portrayed in the figure below move on a frictionless surface and a force F: 36w acts as shown on the 4.0 kg block.
a) Determine the acceleration given this system (in m/s2 to the right). m/s2 (to the right)
b) Determine the tension in the cord connecting the 4.0 kg and the 1.0 kg blocks in N). Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N). N (a) What If How would your answers to parts (a) and (b) of this problem change if the 2.0 kg block was now stacked on top of the 1.0 kg block? Assume that the 2.0 kg block sticks to and does not slide on the 1.0 kg block when the system is accelerated.
(Enter the acceleration in m/s2 to the right and the tension in N.) acceleration m/s (to the right) tension
Answer:
a) 5.143 m/s^2
b) T = 15.43 N
c) Fr = 10.29 N
d) 5.143 m/s^2 , T = 15.43 N
Explanation:
Given:-
- The mass of left most block, m1 = 1.0 kg
- The mass of center block, m2 = 2.0 kg
- The mass of right most block, m3 = 4.0 kg
- A force that acts on the right most block, F = 36 N
Solution:-
a)
- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.
- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:
[tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]
Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2
b)
- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).
- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.
- We will again apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]
Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )
c)
- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).
- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]
- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )
Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.
d)
- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2
- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).
- We apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]
Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.
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.
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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.
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]
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An object will sink in a liquid if the density of the object is greater than that of the liquid. The mass of a sphere is 0.723 g. If the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density
Answer:
= 0.0532 cm^3
Explanation:
The computation of volume of the sphere is shown below:-
[tex]Density = \frac{Mass}{Volume}[/tex]
Where,
Density = 13.6 g/cm^3
Mass of sphere = 0.723 g
now we will put the values into the above formula to reach volume of the sphere which is here below:-
[tex]Volume = \frac{0.723}{13.6}[/tex]
= 0.0532 cm^3
Therefore for computing the volume of the sphere we simply applied the above formula.
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:
https://brainly.com/question/8980142
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Answer: C.D
Explanation:...
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.
A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
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
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
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.
4. Mrs. Parker was married to her husband for
30 years. They lived together with their two
children,
(A) Single-parent family
(B) Nuclear family
(C) Blended family
(D) Extended family
I think it’sd
Explanation:
The answer is B because Nuclear family mean a family with two kids and Mrs. Parker have two kids
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.
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
