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elena-s [515]
3 years ago
5

When light is reflected by a mirror, the angle of an incidence is always?

Physics
1 answer:
Aliun [14]3 years ago
5 0
It's C, angle of incidence is equal to angle of reflection
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How can you safely experiment with plants during a scientific investigation?
liubo4ka [24]

I would think the answer is c.

3 0
3 years ago
Read 2 more answers
A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

4 0
3 years ago
A bullet with a mass ????b=13.5 g is fired into a block of wood at velocity ????b=249. The block is attached to a spring that ha
larisa86 [58]

Answer:

M = 0.436 kg

Explanation:

As per energy conservation we can say that energy stored in the spring at the position of maximum compression must be equal to the kinetic energy of bullet and block system

so here we have

\frac{1}{2}(m + M)v^2 = \frac{1}{2} kx^2

here we know that

k = 205 N/m

x = 35 cm

\frac{1}{2}(m + M)v^2 = \frac{1}{2}(205)(0.35)^2

now by momentum conservation we know that

mv_o = (m + M)v

v = \frac{m}{m + M} v_o

now plug in all values in it

v = \frac{0.0135}{0.0135 + M}(249)

now from above equation

\frac{1}{2}(0.0135 + M)( \frac{0.0135}{0.0135 + M}(249))^2 = 12.56

\frac{5.65}{0.0135 + M} = 12.56

by solving above equation we have

M = 0.436 kg

3 0
3 years ago
An object at rest starts accelerating.
Shalnov [3]

Answer:

<u>We are given: </u>

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

<u>Solving for 'a'</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0
3 years ago
What is the KE of a 27 kg mass moving at 3m/s?
Naily [24]

Answer:

121.5 J

Explanation:

3 0
3 years ago
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