Answer:
Explanation:
400 W = 400 J/s
300000 J / 400 J/s = 750 s or 12.5 minutes
Answer:
Explanation:
According to heisenberg uncertainty Principle
Δx Δp ≥ h / 4π , where Δx is uncertainty in position , Δp is uncertainty in momentum .
Given
Δx = 1 nm
Δp ≥ h /1nm x 4π
≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x 4 π
≥ . 5254 x ⁻²⁵
h / λ ≥ . 5254 x ⁻²⁵
6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ
12.56 x 10⁻⁹ ≥ λ
longest wave length = 12.56 n m
Answer:
THE RUBBER BALL
Explanation:
From the question we are told that
The mass of the rubber ball is ![m_r = 2 \ kg](https://tex.z-dn.net/?f=m_r%20%20%20%3D%20%202%20%5C%20kg)
The initial speed of the rubber ball is ![u = 3 \ m/s](https://tex.z-dn.net/?f=u%20%3D%20%203%20%5C%20m%2Fs)
The final speed at which it bounces bank ![v - 3 \ m/s](https://tex.z-dn.net/?f=v%20%20-%203%20%5C%20m%2Fs)
The mass of the clay ball is ![m_c = 2 \ kg](https://tex.z-dn.net/?f=m_c%20%3D%20%202%20%20%5C%20kg)
The initial speed of the clay ball is ![u = 3 \ m/s](https://tex.z-dn.net/?f=u%20%3D%203%20%5C%20m%2Fs)
The final speed of the clay ball is ![v = 0 \ m/s](https://tex.z-dn.net/?f=v%20%3D%200%20%5C%20%20m%2Fs)
Generally Impulse is mathematically represented as
where
is the change in the linear momentum so
![I = m(v-u)](https://tex.z-dn.net/?f=I%20%20%3D%20%20m%28v-u%29)
For the rubber is
![I_r = 2(-3 -3)](https://tex.z-dn.net/?f=I_r%20%20%3D%20%202%28-3%20-3%29)
![I_r = -12\ kg \cdot m/s](https://tex.z-dn.net/?f=I_r%20%20%3D%20-12%5C%20kg%20%5Ccdot%20%20m%2Fs)
=> ![|I_r| = 12\ kg \cdot m/s](https://tex.z-dn.net/?f=%7CI_r%7C%20%20%3D%2012%5C%20kg%20%5Ccdot%20%20m%2Fs)
For the clay ball
![I_c = 2(0-3)](https://tex.z-dn.net/?f=I_c%20%20%3D%20%202%280-3%29)
![I_c = -6 \ kg\cdot \ m/s](https://tex.z-dn.net/?f=I_c%20%3D%20%20-6%20%5C%20kg%5Ccdot%20%5C%20m%2Fs)
=> ![| I_c| = 6 \ kg\cdot \ m/s](https://tex.z-dn.net/?f=%7C%20I_c%7C%20%3D%20%206%20%5C%20kg%5Ccdot%20%5C%20m%2Fs)
So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball