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2 years ago

If the light strikes the first mirror at an angle θ1, what is the reflected angle θ2? express your answer in terms of θ1.

Physics

1 answer:

Alexus [3.1K]2 years ago

3 0

Answer:

θ₂ = 90° - θ₁

Explanation:

When the light falls on a mirror it bounces back. This is know as reflection. The incident angle is equal to the angle of reflection.

Here, the light strikes the mirror at an angle = θ₁

To find the angle of reflection we first need to understand angle of incidence. The angle of incidence is the angle made between the incident ray and normal. Normal is an imaginary line drawn perpendicular line on the boundary of the mirror.

Since the light strikes the mirror at angle of θ₁, which is the angle between light ray and the mirror.

Angle of incidence = 90° - θ₁.

Thus, angle of reflection, θ₂ = 90° - θ₁

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Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 8325 re

alexgriva [62]

Answer:

$\frac{Ie}{lm} = 1.10*10^{-3}$

Explanation:

GIVEN DATA:

Engine operating speed nf = 8325 rev/min

engine angular speed ni= 12125 rev/min

motorcycle angular speed N_m= - 4.2 rev/min

ratio of moment of inertia of engine to motorcycle is given as

$\frac{Ie}{lm} = \frac{-N}{(nf-ni)}$

$\frac{Ie}{lm} = \frac{-(-4.2)}{(12125 - (8325))}$

$\frac{Ie}{lm} = 1.10*10^{-3}$

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2 years ago

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A hiker walks 200m west and then walks 100m north. What is the magnitude and direction of her resulting displacement?

otez555 [7]

A hiker walks 200 m west and then walks 100 m north, the resulting magnitude is 223 m. The direction can be solve using trigo function sin angle = opposite/ hypotenuse. Which is equal to 26.6 degree. So the displacement and direction is 223 m 26.6 degree North of west.

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2 years ago

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A tennis player who is recovering from an ankle injury and is not allowed to change directions can maintain her cardio fitness l

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A tennis player who is recovering from an ankle injury and is not allowed to change directions can maintain her cardio fitness level by using the rowing machine, the stationary bike with one leg, or swimming. These activities do not require a change in direction. These activities will not give strain to the injured part.

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A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g

Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ( $\varDelta T$ )in the next way:

$Q=cm\varDelta T$ (2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality

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2 years ago

A body covers a semicircle of radius 7cm in 5s .find its linear speed

choli [55]

Ok so we are given the radius of 7cm and time of 5 seconds.

From the data we got we can calculate speed, frequency, perimeter and area of the semicircle.

Let's start with perimeter.

We know that perimeter of circle is $2\pi r$ so the perimeter of semicircle is $\dfrac{2\pi r}{2}$ or simply $\pi r$

So the perimeter is equal to:

$\pi r=\pi\cdot7\approx\boxed{22cm}$

So this is the length of a curve or let's say the distance.

Now let's look at the linear speed $s=\dfrac{d}{t}$ where d is distance and t time.

We know the distance and we know the time.

So let's calculate it.

$s=\dfrac{d}{t}=\dfrac{22}{5}=\boxed{4.4\dfrac{cm}{s}}$

Hope this helps.

r3t40

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2 years ago