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BaLLatris [955]
3 years ago
8

What happens to light when it strikes the inside surface of a smooth, curved mirror?

Physics
2 answers:
soldier1979 [14.2K]3 years ago
7 0

Answer:

The answer is it bounces towards a single spot.

Explanation:

Light bounces off of mirrors, meaning the light is a reflection on a smooth surface. Reflection on smooth surfaces reflect in a single direction.

Zinaida [17]3 years ago
3 0
It bounces in random directions
It bounces back toward a single spot.
It passes through the mirror and moves in random directions.
It passes through the mirror and moves in a straight line.
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A 24 toy falls from 2 to 1 m. How much does the toys CPE change
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GPE change:

\tt =24\times 9.8(2-1)=235.2~J

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two objects were lifted by machine one object had a massive 2 kg it was lifted at a speed of 2 m/s the other had a massive 4 kg
rosijanka [135]
Kinetic energy =(1/2) (mass) (speed²)

First object:  KE = (1/2) (2 kg) (2m/s)² =  4 joules during the lift.

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The second object has more kinetic energy while it's being lifted
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What is the force on object exerts on another?<br>​
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4 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
3 years ago
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