Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
Answer:
Explanation:
Given
![N_1=1 rev/s](https://tex.z-dn.net/?f=N_1%3D1%20rev%2Fs)
angular velocity ![\omega =2\pi N_1=6.284 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D2%5Cpi%20N_1%3D6.284%20rad%2Fs)
Combined moment of inertia of stool,student and bricks ![=6\ kg.m^2](https://tex.z-dn.net/?f=%3D6%5C%20kg.m%5E2)
Now student pull off his hands so as to increase its speed to suppose
rev/s
After Pulling off hands so final moment of inertia is
![I_2=2\ kg-m^2](https://tex.z-dn.net/?f=I_2%3D2%5C%20kg-m%5E2)
Conserving angular momentum as no external torque is applied
![I_1\omega _1=I_2\omega _2](https://tex.z-dn.net/?f=I_1%5Comega%20_1%3DI_2%5Comega%20_2)
![6\times 6.284=2\times \omega _2](https://tex.z-dn.net/?f=6%5Ctimes%206.284%3D2%5Ctimes%20%5Comega%20_2)
![\omega _2=18.85\ rad/s](https://tex.z-dn.net/?f=%5Comega%20_2%3D18.85%5C%20rad%2Fs)
![N_2=3 rev/s](https://tex.z-dn.net/?f=N_2%3D3%20rev%2Fs)
Answer:
![q_2=2.47\times 10^{-4}\ C](https://tex.z-dn.net/?f=q_2%3D2.47%5Ctimes%2010%5E%7B-4%7D%5C%20C)
Explanation:
The charge on one object, ![q_1=9.9\times 10^{-5}\ C](https://tex.z-dn.net/?f=q_1%3D9.9%5Ctimes%2010%5E%7B-5%7D%5C%20C)
The distance between the charges, r = 0.22 m
The force between the charges, F = 4,550 N
Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :
![F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bkq_1q_2%7D%7Br%5E2%7D%5C%5C%5C%5Cq_2%3D%5Cdfrac%7BFr%5E2%7D%7Bkq_1%7D%5C%5C%5C%5Cq_2%3D%5Cdfrac%7B4550%5Ctimes%20%280.22%29%20%5E2%7D%7B9%5Ctimes%2010%5E9%5Ctimes%209.9%5Ctimes%2010%5E%7B-5%7D%7D%5C%5C%5C%5Cq_2%3D2.47%5Ctimes%2010%5E%7B-4%7D%5C%20C)
So, the charge on the other sphere is
.
Answer:
No, you can't keep on dividing the charge forever.
Explanation:
No, you can't keep on dividing the charge in that manner forever because the total charge of the stick is an integer multiples of individual units known as an elementary charge, <em>which is the electron (e) charge (e = 1.602x10⁻¹⁹C)</em>.
Therefore the limit of the division of the original charge will be the electron charge since it is the smallest charge that can exist freely.
I hope it helps you!
Answer:
1.42
Explanation:
<em> got it right on my homework </em>