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3 years ago

What happens when light strikes a translucent object?

Physics

2 answers:

Ne4ueva [31]3 years ago

6 0

Answer:

Explanation:

Its the answer on my usatestprep

sammy [17]3 years ago

3 0

Answer:

D) Some of the light passes through, and some of the light is absorbed or scattered by the object.

Explanation:

When light strikes translucent materials, only some of the light passes through them. The light does not pass directly through the materials. ... When light strikes an opaque object none of it passes through. Most of the light is either reflected by the object or absorbed and converted to heat.

(I googled it) ☺

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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 5.27 m

jeyben [28]

Answer:

See explanation

Explanation:

From Newton's law; the rate of change of momentum is proportional to the impressed force. Hence;

F.t = mv- mu

F= force

t= time

m= mass

V= final velocity

u = initial velocity

since u = 0, mv= 0

F= MV/t

F= 69.8 × 5.27/0.833

F= 442 N

Impulse = Ft= 442 × 0.833= 368 Ns

5 0

2 years ago

Which is TRUE about static electricity?

Fed [463]

Answer:

the first one stationary charge

3 0

2 years ago

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If all blocks are the same density, are they all the same material?

schepotkina [342]

No they are not the same material because having the same density doesnt mean its the same material also they can have a different density and be the same material

8 0

3 years ago

Read 2 more answers

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

3 years ago

Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car

juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

$a = \frac{-F}{m}$

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion

5 0

3 years ago