Answer:
Light behaves differently in space than on Earth.
Explanation:
Because the gravity field is greater near earth than in most of space. Not the areas near stars, black holes, pulsars, and such but in the vast emptyness between the clumpy spots.
Calculate the electric field at point A, located at coordinates (0 m, 12.0 m ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.
Answer:
The correct answer is "[tex](0,300\times 10^{-3} \ N/C)[/tex]".
Explanation:
The given problem seems to be incomplete. Please find the attachment of the complete query.
According to the question,
At point A, we have
⇒ [tex]E_x = \frac{k q_1}{d_1^2} Cos \theta_1 - \frac{k q_2}{d_2^2} Cos \theta_2[/tex]
or,
⇒ [tex]E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ][/tex]
[tex]=0[/tex]
and,
⇒ [tex]E_y = \frac{kq_1}{d_1^2}Sin \theta_1 +\frac{kq_2}{d_2^2}Sin \theta_2[/tex]
or,
⇒ [tex]E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ][/tex]
[tex]=0.3 \ N/C[/tex]
A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.5 m/s2
Answer:
Explanation:
Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is
F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes
[tex]F_n-w=ma[/tex] where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:
[tex]F_n=ma+w[/tex] . m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:
w = mg so
w = 28(9.8) and
w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually
w = 270 N.
Filling in the elevator equation:
[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:
[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:
[tex]F_n=280N[/tex] So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?
Calculate the elastic energy stored up in a wire originally 5 meter
long and 10^-3 m in diameter which has been stretched by 3×10^-4 m due to a load of 10 kg.
Answer:
The elastic energy is 245 J.
Explanation:
Length, L = 5 m
Diameter, D = 10^-3 m
Stretch, l = 3 x 10^-4 m
Load, F = 10 x 9.8 = 98 N
Let the elastic energy is U.
[tex]U = \frac{1}{2}\times stress\times strain\times volume\\\\U = 0.5\times \frac{Force}{area}\times \frac{l}{L}\times Area\times L\\\\U = 0.5 \times F\times l\\\\U = 0.5\times 98\times 5\\\\U = 245 J[/tex]
A 5.85-mm-high firefly sits on the axis of, and 13.7 cm in front of, the thin lens A, whose focal length is 5.01 cm. Behind lens A there is another thin lens, lens B, with focal length 25.9 cm. The two lenses share a common axis and are 62.5 cm apart. 1. Is the image of the firefly that lens B forms real or virtual?
a. Real
b. Vrtual
2. How far from lens B is this image located (expressed as a positive number)?
3. What is the height of this image (as a positive number)?
4. Is this image upright or inverted with respect to the firefly?
a. Upright
b. Inverted
Answer:
1. The image is real
2. 5.85
3. h' = 3.05 mm
4. The image is upright
Explanation:
1. Start with the first lens and apply 1/f = 1/p + 1/q
1/5.01 = 1/13.7 + 1/q
q = 7.90 cm
Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,
1/25.9 = 1/54.6 + 1/q
q = 49.3 cm behind the second lens
Using that information, since q is positive, the image is real
2. Also, using that information, you have the second answer, which is 49.3 cm
The height can be found from the two magnifications.
m = -q/p
m1 = -7.9/13.7 = -.577
m2 = -49.3/54.6 = -.903
Net m = (-.577)(-.903) = .521
Then, m = h'/h
.521 = h'/5.85
3. h' = 3.05 mm
4. For the fourth answer, since the overall magnification is positive, the final image is upright
Two loudspeakers, 5.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m . Assume the speed of sound is 340 m/s.
Required:
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.21 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
c.
Solution :
Let [tex]$d_1=\frac{5.5}{2}[/tex]
= 2.75 m
[tex]d_2 = 0.21 \ m[/tex]
And [tex]$d=|d_1-d_2|$[/tex]
[tex]$d=(d_1+d_2) - (d_1-d_2)$[/tex]
[tex]$d=(2.75+0.21) - (2.75-0.21)$[/tex]
[tex]$d = 2.96-2.54$[/tex]
[tex]d = 0.42 \ m[/tex]
a). At minimum,
[tex]$d=\frac{\lambda}{2}$[/tex]
[tex]$\lambda = 2d$[/tex]
= 2 x 0.42
= 0.84 m
Frequency, [tex]$\nu = \frac{v}{\lambda}$[/tex]
[tex]$=\frac{340}{0.84}$[/tex]
= 404.76 Hz
Therefore, the frequency of he sound, [tex]$\nu$[/tex] = 404.76 Hz
b). At maximum, λ = d = 0.42 m
Therefore, the frequency, [tex]$\nu = \frac{v}{\lambda}[/tex]
[tex]$=\frac{350}{0.42}$[/tex]
= 809.52 Hz
27. The part of the Earth where life exists .
Mesosphere
Stratosphere
Troposphere
Biosphere
Answer:
Biosphere is the part of the earth where life exists.
A student of mass 50kg takes 15seconds to run up a flight of 50 steps. If each step is 20cm, calculate the potential energy of the student at the maximum height
Answer:
the answer is 49000 joules at the maximum height
Explanation:
we know the mass (50kg)
we know the acceleration due to gravity(9.8m/s²)
we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m
the formula is potential energy=mgh
m for mass
g for acceleration due to gravity
h for height
multiply them
50x9.8x100
we get 49000
the unit of potential energy is joules so the answer is
49000 joules
Answer:
49000 joules
Explanation:
hope it helpss
In order to test an intentionally weak adhesive, the bottom of the small 0.15-lb block is coated with adhesive and then the block is pressed onto the turntable with a known force. The turntable starts from rest at t = 0 and uniformly accelerates with a = 2 rad/s^2. If the adhesive fails at exactly t = 3 s, then determine:
a. the magnitude of the ultimate shear force that the adhesive supports
b. the angular displacement of the turntable at the time of failure
Answer:
answer
Explanation:
it is the answer which was presented in the year
the lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
radius of a wire is 1.8mmm,Use value of resistivity for Cu as 1.75 × 10-8Ωm.
Answer:
R = 0.0015Ω
Explanation:
The formula for calculating the resistivity of a material is expressed as;
ρ = RA/l
R is the resistance
ρ is the resistivity
A is the area of the wire
l is the length of the wire
Given
l = 85cm = 0.85m
A = πr²
A = 3.14*0.0018²
A = 0.0000101736m²
ρ = 1.75 × 10-8Ωm.
Substitute into the formula
1.75 × 10-8 = 0.0000101736R/0.85
1.4875× 10-8 = 0.0000101736R
R = 1.4875× 10-8/0.0000101736
R = 0.0015Ω
A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front wheel on the bike is 750 g and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the bike leaves the ground?
Answer:
Explanation:
The formula for angular momentum is
L = mvr where L is the angular momentum, m is the mass of the object, v is the velocity of the object, and r is the radius of the object. The problem we have that prevents us from just throwing those numbers in there is that mass has to be in kg and it's not, and radius has to be in meters and it's not.
Changing the mass to kg:
750 g = .750 kg
Changing the radius to m:
35 cm = .35 m
Now we can fill in the variables with their respective values:
L = .750(10.0)(.35) gives us
[tex]L=2.625\frac{kg*m^2}{s}[/tex]
Thorium-232 goes through multiple types of decay in order to reach a stable isotope. What isotope is created after the first two decays if it first goes through an alpha decay and then a beta decay?
A)uranium-236
B)protactinium-232
C)radon-224
D)Astinium-228
Answer:
The answer would be D), if the decay is beta negative.
Explanation:
Thorium-232 goes through alpha decay:
Thorium-232 --> Helium-4 + Radium-228
Radium-228 then can undergo beta positive or beta negative decay:
Beta positive = Radium-228 --> Electron + Francium-228
Beta negative = Radium-228 --> Positron + Actinium-228
Therefore, the isotope that is created is Actinium-228
Si un resorte de constante elástica 1300 n/m se comprime 12 cm ¿Cuanta energía almacena? Y si estira 12cm ¿Cuanta energía almacena?
La energía que almacena el resorte cuando se comprime y estira 12 cm es 9,4 J.
La energía potencial elástica del resorte se puede calcular con la siguiente ecuación:
[tex] E_{p} = \frac{1}{2}kx^{2} [/tex]
En donde:
k: es la constante del resorte = 1300 N/m
x: es la distancia de compresión o de elongación = 12 cm = 0,12 m
Dado que la energía es proporcional al cuadrado de la distancia recorrida por el resorte (x), la energía almacenada por el resorte durante la compresión será la misma que la energía almacenada por la elongación.
Por lo tanto, la energía almacenada es:
[tex]E_{p} = \frac{1}{2}kx^{2} = \frac{1}{2}1300 N/m*(0,12 m)^{2} = 9,4 J[/tex]
Entonces, la energía del resorte cuando se comprime y cuando se estira es la misma, a saber 9,4 J.
Para saber más sobre energía potencial visita este link: https://brainly.com/question/156316?referrer=searchResults
Espero que te sea de utilidad!
Answer:
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
Explanation:
La Energía Potencial Elástica almacenada por el resorte ([tex]U_{e}[/tex]), en joules, se calcula a partir de la Ley de Hooke, la definición de Trabajo y el Teorema del Trabajo y la Energía, cuya expresión se presenta abajo:
[tex]U_{e} = \frac{1}{2}\cdot k\cdot (x_{f}^{2}-x_{o}^{2})[/tex] (1)
Donde:
[tex]k[/tex] - Constante elástica del resorte, en newtons por metro.
[tex]x_{o}[/tex] - Posición inicial del resorte, en metros.
[tex]x_{f}[/tex] - Posición final del resorte, en metros.
Nótese que el resorte sin deformar tiene una posición de cero, la tensión tiene un valor positivo y la compresión, negativo.
Asumiendo que en ambos casos el resorte se encuentra inicialmente sin deformar, se reduce (1) a una forma de función par, es decir, una función que cumple con la propiedad de que [tex]f(x) = f(-x)[/tex], se encuentra que al comprimirse o estirarse en la misma medida almacena la misma cantidad de energía.
La cantidad de energía a almacenar es:
[tex]U_{e} = \frac{1}{2}\cdot \left(1300\,\frac{N}{m} \right)\cdot (0,12\,m)^{2}[/tex]
[tex]U_{e} = 9,360\,J[/tex]
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
Particle A has less mass than particle B. Both are pushed forward across a frictionless surface by equal forces for 1 s. Both start from rest. Which is true? A. A has more momentum. B. B has more momentum. C. A and B have the same momentum D. Not enough information.
Answer:
Both will have the same momentum.
P = M v momentum
v = a t for uniform acceleration
P = M a t
But a = F / M
P = M (F / M) t = F t so both have the same momentum
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
Physics question plz help ASAP
Calculate the potential energy stored in a metal ball of a mass of 80 kg kept at a height of 15m from the earth surface.What will be the potential energy when the metal ball is kept on the earth surface.
Answer:
39200 joules
the potential energy will be zero
Explanation:
we know that potential energy is found by multiplying mass, acceleration due to gravity and height from the Earth's surface
so it will be
potential energy= mgh
80x9.8x15
= 39200 joules
the potential energy of the mental ball will be zero when kept on the Earth's surface because the height from the Earth's surface will be zero and zero multiplied to any number is zero only
I have a doubt with the second one, this is what I think it is. Consult your teacher if you think my answer for the second one is wrong
Answer:
392000 joules
Explanation:
hope it helpsss
write down the unit of mass ,temperature ,power and density
Explanation:
mass=kilogram,temperature=Klevin,power=watt,density=kilogram per cubic metre
Explanation:
the unit of mass is kg , temperature is kelvin ,power is watt and density is kilogram per cubic meter.
Which of these rotational quantities is analogous to mass in a linear system?
a.
Angle in radians
b.
Angular acceleration
c.
Torque
d.
Rotational inertia
Answer:
d
Explanation:
i think it is rotational inertia
because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.
hope it's right & helps !!!!!!!!!
A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.
Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.
The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.
Momentum is conserved, so the total momentum of the system is the same before and after the collision:
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
==>
(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'
==>
-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'
where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.
Kinetic energy is also conserved, so that
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²
==>
(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
==>
1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is
v₁' ≈ -3.11 m/s
v₂' ≈ -0.167 m/s
and take the absolute values to get the magnitudes.
If you want to instead use the masses from the "Required" section, you would end up with
v₁' ≈ -3.18 m/s
v₂' ≈ -0.236 m/s
Which of the following has a negative acceleration?
A. A car increases its speed moving forward.
B. A car sits at rest at a stop sign.
C. A car is slowing down as it approaches a traffic light.
D. A car is in cruise control at a constant speed.
Answer:
B. A car sits at rest at a stop sign.
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound
The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]
Therefore,
[tex]$AD-BD = \frac{\lambda}{2}[/tex]
[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]
[tex]$\lambda = 2 \times 0.4985$[/tex]
[tex]$\lambda = 0.99714 \ m$[/tex]
Thus the frequency is :
[tex]$f=\frac{v}{\lambda}$[/tex]
[tex]$f=\frac{340}{0.99714}$[/tex]
[tex]f=340.9744[/tex] Hz
A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in diameter and has 1500 turns. When turned on, the current in the solenoid is increases linearly to 20 A in 1 second. What is the induced emf in the ring?
a) 2.0 x 10-5 v
b) 3.8 x 10-5 v
c) 1.2 x 10-3 v
d) 1.9 x 10-4 v
Answer:
the answer should be b) 3.8 x 10-5 v
Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is Q=ΔV/Δt what is the flow rate in pipe A? Viscosity: Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is what is the flow rate in pipe A?
a) Q√2
b) 16Q
c) 2Q
d) 4Q
e) 8Q
Answer:
c) 2Q
Explanation:
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D = D_A = D_B[/tex]
where;
length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]
[tex]\mathbf{Q_1 = 2Q}[/tex]
The flow rate in pipe B is 2Q of the flow rate of the pipe A
What is flow rate?
The flow rate is defined as the flow of the fluid across the cross section in per unit time.
From the given information:
The pressure inside a pipe can be expressed by using the formula:
[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]
Since the diameter in both pipes is the same, we can say:
[tex]D=D_A=D_B[/tex]
where;
length of the first pipe A [tex]L_A=L[/tex] and the length of the second pipe B
[tex]L_B=2L[/tex]
Since the difference in pressure is equivalent in both pipes:
Then:
[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]
[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]
[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]
[tex]Q_1=2Q[/tex]
Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A
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A simple pendulum consists of a ball of mass 3 kg hanging from a uniform string of mass 0.05 kg and length L. If the period of oscillation of the pendulum is 2 s, determine the speed of a transverse wave in the string when the pendulum hangs vertically.
Answer:
v = 3.12 m/s
Explanation:
First, we will find the length of the string by using the formula of the time period:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\l = \frac{T^2g}{4\pi^2}\\\\[/tex]
where,
l = length of string = ?
T = time period = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]l = \frac{(2\ s)^2(9.81\ m/s^2)}{4\pi^2}\\\\l = 0.99\ m[/tex]
Now, we will find tension in the string in the vertical position through the weight of the ball:
T = W = mg = (3 kg)(9.81 m/s²)
T = 29.43 N
Now, the speed of the transverse wave is given as follows:
[tex]v=\sqrt{\frac{Tl}{m}}\\\\v=\sqrt{\frac{(29.43\ N)(0.99\ m)}{3\ kg}}\\\\[/tex]
v = 3.12 m/s
A force of 200 N, acting at 60° to the horizontal, accelerates a block of mass 50 kg along a horizontal plane. Calculate the component of the 200N force that accelerates the block horizontally
Answer:
Explanation:
a) Fx = F cos (θ)
= (200) cos(60)
= 100 N
b) FR = ma
Fx + Ff = ma
100 + Ff = (50)(1,5)
Ff = 75 - 100
= -25 N
c) Fy = F sin θ
= (200) sin(60)
= 173,2 N
Light falling on a metal surface causes electrons to be emitted from the metal by the photoelectric effect.
As we decrease the frequency of this light, but do not vary anything else (there may be more than one correct answer),
A: the number of electrons emitted from the metal increases.
B: the maximum speed of the emitted electrons decreases.
C: the maximum speed of the emitted electrons does not change.
D: the work function of the metal increases.
(B)
Explanation:
The speed of the ejected electrons depends on the frequency of the incident radiation. The closer the energy of the incident photons to the work function of the metal, the slower is the speed of the ejected electrons. Intensity of the incident radiation has no effect on the speed of the ejected electrons, only its frequency.
Light falling on a metal surface causes electrons to be emitted from the metal by the photoelectric effect. As we decrease the frequency of this light, but do not vary anything else B: the maximum speed of the emitted electrons decreases.
What is photoelectric effect?
Photoelectric effect is the phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.
According to Photoelectric effect the kinetic energy of the photo electrons emitted depend on the frequency of incident light , the more is the frequency the more is the kinetic energy of emitted electron and hence high will be the velocity of the emitted electrons and vise versa
since , in question the frequency has been decreased hence , the kinetic energy must be decreased therefore velocity will also get decreased
hence , correct option will be B: the maximum speed of the emitted electrons decreases.
learn more about Photoelectric effect
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What computer measures physical quantities?
Answer:
Three Types of Computer The Computer are classified into three main types:: • Analog Computers • Digital Computers • Hybrid Computers (Analog + Digital) Analog Computers:: Analog Computer measures “Physical Quantities” for example Temperature, Voltage, Pressure, and Electric Current.
Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )
1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )
1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) = [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ = [tex]\frac{V}{i+0.5}[/tex]
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ = [tex]\frac{V}{i+0.25}[/tex]
R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]
R₂ = 19.9 Ω
An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions. I live 10 km from this station. What is the maximum strength of Electric Field in my house
Answer:
[tex]E_0=0.173N/C[/tex]
Explanation:
From the question we are told that:
Power [tex]P=50kw=>50*10^3w[/tex]
Distance [tex]d=10km=10000m[/tex]
Generally the equation for Intensity is mathematically given by
[tex]I=\frac{P}{4\pi d^2} w/m^2[/tex]
[tex]I=\frac{50*10^3}{4 \pi 10000^2} w/m^2[/tex]
[tex]I=3.98*10^{-5}w/m^2[/tex]
Generally Intensity is also
[tex]I=\frac{1}{2}cE_0^2e[/tex]
Where
[tex]e=8.854*10^{-12}Nm^2/c^2[/tex]
Therefore
[tex]E_0=\sqrt{\frac{2I}{c *e}}[/tex]
[tex]E_0=\sqrt{\frac{2*3.98*10^{-5}}{3*10^8 *8.854*10^{-12}}}[/tex]
[tex]E_0=0.173N/C[/tex]
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
