If you cannot get a chair to move across the floor, it is because static friction opposes your push. When you say static or kinetic friction the two object that facing each other are opposing each other. That's why you're having a hard time pushing the chair.
Answer:
t = 2.58*10^-6 s
Explanation:
For a nonconducting sphere you have that the value of the electric field, depends of the region:
![rR:\\\\E=k\frac{Q}{r^2}](https://tex.z-dn.net/?f=r%3CR%3A%5C%5C%5C%5CE%3Dk%5Cfrac%7BQr%7D%7BR%5E3%7D%5C%5C%5C%5Cr%3ER%3A%5C%5C%5C%5CE%3Dk%5Cfrac%7BQ%7D%7Br%5E2%7D)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the sphere = 10.0/2 = 5.0cm=0.005m
In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:
![F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}](https://tex.z-dn.net/?f=F%3Dm_pa%5C%5C%5C%5CqE%3Dm_pa%5C%5C%5C%5Ck%5Cfrac%7BqQ%7D%7Br%5E2%7D%3Dm_pa%5C%5C%5C%5Ca%3Dk%5Cfrac%7BqQ%7D%7Bm_pr%5E2%7D)
Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:
![Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}](https://tex.z-dn.net/?f=Q%3D%5Crho%20V%3D%285.5%2A10%5E%7B-6%7DC%2Fm%5E3%29%28%5Cfrac%7B4%7D%7B3%7D%5Cpi%280.05m%29%5E3%29%3D2.87%2A10%5E%7B-9%7DC%5C%5C%5C%5Ca%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5Cfrac%7B%281.6%2A10%5E%7B-19%7DC%29%282.87%2A10%5E%7B-9%7DC%29%7D%7B%281.67%2A10%5E%7B-27%7Dkg%29%280.05m%29%5E2%7D%3D9.87%2A10%5E%7B11%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D)
with this values of a you can use the following formula:
![a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-v_o%7D%7Bt%7D%5C%5C%5C%5Ct%3D%5Cfrac%7Bv-v_o%7D%7Ba%7D%3D%5Cfrac%7B2550%2A10%5E3m%2Fs-0m%2Fs%7D%7B9.87%2A10%5E%7B11%7Dm%2Fs%5E2%7D%3D2.58%2A10%5E%7B-6%7Ds)
hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s
Answer:
a. A uniform disk of radius and mass .
Explanation:
The moment of inertia I of an object depends on a chosen axis and the mass of the object. Given the axis through the point, the inertia will be drawn from the uniform disc having a radius and the mass.
.
Power is the amount of work done over a period of time. If you will put that into an equation, the formula of power will be:
P = W/t
Where:
P = power
W=work
t = time
Your problem already provides you with work and time so all you need to do is divide:
P = W/t
P = 50J/30s
P = 1.67 W
Answer:
75.36 mph
Explanation:
The distance between the other car and the intersection is,
The distance between the police car and the intersection is,
(Negative sign indicates that he is moving towards the intersection)
Therefore the distance between them is given by,
The rate of change is,
Now finding
when
from (1) we have
The officer's radar gun indicates 25 mph pointed at the other car then,
when
from
From (2) we get
Hence the speed of the car is ![75.36 mph](https://tex.z-dn.net/?f=75.36%20mph)