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3 years ago

How does a mirror affect the path of light

Physics

1 answer:

Slav-nsk [51]3 years ago

5 0

Answer:

light bounces off the mirror

Explanation:

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Can you change your mass without changing your weight?

Ratling [72]

No, because mass is the amount of matter in something and weight is the pull of gravity on a object, for example you would weigh 65 pounds.. and have the mass of 1058..if you were in space your weight would change but your mass will always stay the same

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4 years ago

At time t0 = 0 the mass happens to be at y0 = 4.05 cm and moving upward at velocity v0 = +4.12 m/s. (Mind the units!) Calculate

Katen [24]

Answer:

A = 5.727 cm

Explanation:

at time t = 0s the displacement of mass is 4.05 cm and velocity 4.12 m/s

we know that velcoity for simple harmonic motion is given as

$v_o = \omega \sqrt {A^2 - Y_O^2}$

W KNOW THAT

$\omega = \frac{v_0}{Y_o}$

Therefore we get

$Y_O^2 = A^2 - Y_O^2$

$A =\sqrt 2 * Y_O$

$A =\sqrt 2 * 4.05\\$

A = 5.727 cm

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3 years ago

Two extremely large nonconducting horizontal sheets each carry uniform charge density on the surfaces facing each other. The upp

sashaice [31]

This problem refers to a parallel plate capacitor. There is an electric field between the two plates. The working equation to be used is the Gauss’s Law which is

Electric field = Surface charge density / ε0

The answer is -2.52 μC/m2.

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3 years ago

What is the current flowing through a 10 resistor in a parallel circuit with a 120 V potential difference across it?

Elena L [17]

Answer:

let current flow be x

Explanation:

than we know potential difference = energy supply / charge then 120 = 10 /x then x=12 therefore pd = 12 v

5 0

3 years ago

Read 2 more answers

A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei

Mariana [72]

Answer:

the speed of the car at the top of the vertical loop $v_{top} = 2.0 \sqrt{gR \ \ }$

the magnitude of the normal force acting on the car at the top of the vertical loop $F_{N} = 1.47 \ \ N$

Explanation:

Using the law of conservation of energy ;

$mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R$

$v_{top} = 2.0 \sqrt{gR \ \ }$

The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

$F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\$

$F_{N} = 1.47 \ \ N$

4 0

4 years ago