Answer:
<em>c. tension, gravity, and the centripetal force</em>
<em></em>
Explanation:
The ball experiences a variety of force as explained below.
Gravity force acts on the body due to its mass and the acceleration due to gravity. The gravity force on every object on earth due to its mass keeps all object on the surface of the earth.
Although the car moves around in circle, centripetal towards the center of the radius of turn exists on the ball. This centripetal force is due to the constantly changing direction of the circular motion, resulting in a force away from the center. The centripetal force keeps the ball from swinging off away from the center of turn.
Tension force on the string holds the ball against falling towards the earth under its own weight, and also from swinging away from the center of turn of the car. Tension force holds the ball relatively fixed in its vertical position in the car.
D) There will be less available for other living things.
Answer:
<h2>160 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
mass = 80 kg
acceleration = 2 m/s²
We have
force = 80 × 2 = 160
We have the final answer as
<h3>160 N</h3>
Hope this helps you
Answer:
![K=512J](https://tex.z-dn.net/?f=K%3D512J)
Explanation:
Since the surface is frictionless, momentum will be conserved. If the bullet of mass
has an initial velocity
and a final velocity
and the block of mass
has an initial velocity
and a final velocity
then the initial and final momentum of the system will be:
![p_i=m_1v_{1i}+m_2v_{2i}](https://tex.z-dn.net/?f=p_i%3Dm_1v_%7B1i%7D%2Bm_2v_%7B2i%7D)
![p_f=m_1v_{1f}+m_2v_{2f}](https://tex.z-dn.net/?f=p_f%3Dm_1v_%7B1f%7D%2Bm_2v_%7B2f%7D)
Since momentum is conserved,
, which means:
![m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}](https://tex.z-dn.net/?f=m_1v_%7B1i%7D%2Bm_2v_%7B2i%7D%3Dm_1v_%7B1f%7D%2Bm_2v_%7B2f%7D)
We know that the block is brought to rest by the collision, which means
and leaves us with:
![m_1v_{1i}+m_2v_{2i}=m_1v_{1f}](https://tex.z-dn.net/?f=m_1v_%7B1i%7D%2Bm_2v_%7B2i%7D%3Dm_1v_%7B1f%7D)
which is the same as:
![v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}](https://tex.z-dn.net/?f=v_%7B1f%7D%3D%5Cfrac%7Bm_1v_%7B1i%7D%2Bm_2v_%7B2i%7D%7D%7Bm_1%7D)
Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:
![v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s](https://tex.z-dn.net/?f=v_%7B1f%7D%3D%5Cfrac%7B%280.01kg%29%282000m%2Fs%29%2B%284kg%29%28-4.2m%2Fs%29%7D%7B0.01kg%7D%3D320m%2Fs)
So kinetic energy of the bullet as it emerges from the block will be:
![K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7Bmv%5E2%7D%7B2%7D%3D%5Cfrac%7B%280.01kg%29%28320m%2Fs%29%5E2%7D%7B2%7D%3D512J)
Note: I assume the beam comes from air (refractive index: n=1.00).
We can solve the problem by using Snell's law, which states:
![n_i \sin \theta_i = n_r \sin \theta_r](https://tex.z-dn.net/?f=n_i%20%5Csin%20%5Ctheta_i%20%3D%20n_r%20%5Csin%20%5Ctheta_r)
where
![n_i](https://tex.z-dn.net/?f=%20n_i)
is the refractive index of the first medium
![\theta_i](https://tex.z-dn.net/?f=%5Ctheta_i%20)
is the angle of incidence (measured with respect to the normal to the surface)
![n_r](https://tex.z-dn.net/?f=n_r%20)
is the refractive index of the second medium
![\theta_r](https://tex.z-dn.net/?f=%5Ctheta_r)
is the angle of refraction (measured with respect to the normal to the surface).
In our problem, we have
![n_i = 1.00](https://tex.z-dn.net/?f=n_i%20%3D%201.00)
![n_r = 1.58](https://tex.z-dn.net/?f=n_r%20%3D%201.58)
![\theta_i = 46^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_i%20%3D%2046%5E%7B%5Ccirc%7D)
Therefore we can re-arrange Snell's law to find the angle of refraction: