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3 years ago

What happens when light bounces off an object

2 answers:

7 0

When light bounces off an object, it either gets absorbed, or reflected. For example, light can get absorbed on surfaces which are black in color and they can reflect from flat surfaces such as a mirror. It all matters on the angle of incidence. The angle of incidence is also known as the angle of reflection.

Angle of reflection = Angle of incidence

The ray of light approaching the mirror is called the incident ray. The ray of light that leaves the mirror is called the reflected ray.

Hope it helps

Sever21 [200]3 years ago

3 0

Absorbtion consists of when light strikes on an object and bounces off.

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Bonnie and Clyde are trying to steal the world's largest diamond from a 10 story

german

Answer:

The speed Clyde will be falling at is 33.72.

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3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

3 years ago

Look at the figure, which shows the motion of three balls. The curved paths followed by balls B and C are examples of _____. pro

Natali [406]

Answer:did you get the answer?

Explanation:

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3 years ago

Read 2 more answers

A rope pulls a 82.5 kg skier at a constant speed up a 18.7Â° slope with Î¼k = 0.150. How much force does the rope exert?

Artist 52 [7]

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f = μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

6 0

3 years ago

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

3 years ago