Answer:
-A screw is an inclined plane wrapped around a cylinder.
Explanation:
Answer:

Explanation:
Given data
Ball one
mass m₁=3.0kg
velocity v₁=20 m/s
Ball second
mass m₂=2.0 kg
velocity v₂=12 m/s
First we need the speed of combined ball.Since the system conserves the linear momentum

So the combined velocity vt is:

Since the two balls 1 and 2 are moving in opposite direction
So

Substitute the given values

We have the equation for motion with constant acceleration is given by:

At initial position y₀=0 and vt=v-v₀
So

Answer:

Explanation:
The expression which represent the first diffraction minima by a circular aperture is given by
--------eqn 1
The angle through which the first minima is diffracted is given by
---------eqn 2
As
is very small so we can write 
So from eqn 1 and eqn 2 we can write
--------eqn 3
Here
is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture
It is given that diameter of circular aperture is 14.7 cm so 
Now putting all these value in eqn 3


Answer:
The radius is 
Explanation:
From the question we are told that
The distance beneath the liquid is 
The refractive index of the liquid is 
Now the critical value is mathematically represented as
![\theta = sin ^{-1} [\frac{1}{n_i} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7Bn_i%7D%20%5D)
substituting values
![\theta = sin ^{-1} [\frac{1}{131} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B131%7D%20%5D)

Using SOHCAHTOA rule we have that

=> 
substituting values


Answer:
The correct option is;
C. 1,715 m
Explanation:
We are given the information from the group of teen at the City edge
Time of arrival of explosion sound = 5 s after sighting
Time of sighting explosion = 5 s before hearing the boom
Speed of sound in air ≈ 343 m/s
Speed of light = 299,792 km/s
Therefore, distance covered by sound in 5 seconds is given by the following equation;


Hence Distance = 343 m/s × 5 s = 1715 m
To check, we compare the time it would take for the light to cover 1715 m
That is
which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.
Therefore, the distance of the students from the factory is approximately 1,715 m