Ask question

Ask question

All categories

2 years ago

13

g Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an an

gle of 55.7 degrees with respect to the normal to the surface. What is the angle of reflection from the second surface

1 answer:

Mamont248 [21]2 years ago

5 0

Answer:

add the angles

HOPE you understand

You might be interested in

What is the answer to this?

jeka57 [31]

Answer:

I think it is the forth one

7 0

2 years ago

Read 2 more answers

A 14.8-g piece of Styrofoam carries a net charge of -0.742 µC and is suspended in equilibrium above the center of a large, horiz

MAXImum [283]

Base on your question where a 14.8g of piece of Styrofoam carries a net charge of -0.742C and is suspended in equilibrium above the center of a large, horizontal sheet of plastic so the ask of the problem is to calculate the charge per unit area on the plastic sheet. The answer would be 21.96

3 0

3 years ago

The nucleus of an atom is composed of unchanged particles as well as of the following

Ivan

I suppose that you wanted write "uncharged". The particles without electrical charge present in the nucleus are called neutrons.

7 0

2 years ago

____ is a pesticide that influenced society by increasing crop yield, but it also caused human and animal sickness.

Ilya [14]

I think your answer would be DDT, but don't hold me to it

5 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

2 years ago