Answer: In the planet there is life and the star is a store of energy where the planet gets its light, life and energy.
Explanation:
Answer:
Mass of the box = 0.9433 kg
Explanation:
Mass of racket-ball
= 0.00427 kg
Velocity of racket-ball before collision
= 22.3 m/s
Velocity of racket-ball after collision with box
= -11.5 m/s
[Since ball is bouncing back, so velocity is taken negative.]
Velocity of the box before collision
= 0 m/s
<em>[Since the box is stationary, so velocity is taken zero]</em>
Velocity of box moving forward after collision
= 1.53 m/s
To find the mas of the box
.
By law of conservation of momentum we have:
Momentum before collision = Momentum after collision
This can be written as:
![p_i=p_f](https://tex.z-dn.net/?f=p_i%3Dp_f)
![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 can plugin the given value to find ![m_2](https://tex.z-dn.net/?f=m_2)
![(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)](https://tex.z-dn.net/?f=%280.0427%5Ctimes%2022.3%29%2B%28m_2%5Ctimes%200%29%3D%280.0427%5Ctimes%20%28-11.5%29%29%28m_2%5Ctimes%201.53%29)
![0.9522+0=-0.4911+1.53m_2](https://tex.z-dn.net/?f=0.9522%2B0%3D-0.4911%2B1.53m_2)
Adding both sides by 0.4911
![0.9522+0.4911=-0.4911+0.4911+1.53m_2](https://tex.z-dn.net/?f=0.9522%2B0.4911%3D-0.4911%2B0.4911%2B1.53m_2)
![1.4433=1.53m_2](https://tex.z-dn.net/?f=1.4433%3D1.53m_2)
Dividing both sides by 1.53.
![\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}](https://tex.z-dn.net/?f=%5Cfrac%7B1.4433%7D%7B1.53%7D%3D%5Cfrac%7B1.53m_2%7D%7B1.53%7D)
![0.9433=m_2](https://tex.z-dn.net/?f=0.9433%3Dm_2)
∴
kg
Mass of the box = 0.9433 kg (Answer)
Answer:
Sample answer: The mass of the car and the speed of the car (determined by the height of the hill) determine whether the car will break the egg.
Answer:
Explanation:
- given S = distance from the first = 3.20cm = 0.032m, t = 1.30×10−8 s
- acceleration = 0.032 X 2 /(1.30×10−8)^2
a = 3.79 x 10^14m/s^2
E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19
E = magnitude of this electric field. = 2156.3N/C
b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as
= 2 X 3.79 x 10^14 X 0.032
= 4.92 X 10^6m/s