We are given
V = 100 V
R = 500 Ω
C = 10^4 F
q(0) = 0
We are asked to find the charge q(t) of the RC capacitor.
To solve this, we use the formula:
q(t) = CV [1 - e^(-t/RC) ]
Substituting thet given values
q(t) = 10^4 (100) [1 - e^(-t/500(10^4) ]
q(t) = 1 x 20^6 [1 - e^(-2x10^7 t) ]
Since we only asked to get the charge q(t) in terms of t, the answer is
q(t) = 1 x 20^6 [1 - e^(-2x10^7 t) ]
Answer:
6 rad/s
Explanation:
In a spring the angular frequency is calculated as follows:
![\omega=\sqrt{\frac{k}{m} }](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D)
where
is the angular frequency,
is the mass of the object in this case
, and
is the constant of the spring.
To calculate the angular frequency, first we need to find the constant
which is calculated as follows:
![k=\frac{F}{x}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7BF%7D%7Bx%7D)
Where
is the force:
, and
is the distance from the equilibrium position:
.
Thus the spring constant:
![k=\frac{18N}{0.25m}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B18N%7D%7B0.25m%7D)
![k=72N/m](https://tex.z-dn.net/?f=k%3D72N%2Fm)
And now we do have everything necessary to calculate the angular frequency:
![\omega=\sqrt{\frac{k}{m} }=\sqrt{\frac{72N/m}{2kg} }=\sqrt{36}](https://tex.z-dn.net/?f=%5Comega%3D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%3D%5Csqrt%7B%5Cfrac%7B72N%2Fm%7D%7B2kg%7D%20%7D%3D%5Csqrt%7B36%7D)
![\omega=6rad/s](https://tex.z-dn.net/?f=%5Comega%3D6rad%2Fs)
the angular frequency of the oscillation is 6 rad/s
Answer:
Part B)
v = 4.98 m/s
Explanation:
Part a)
As the ball is rolling on the inclined the the friction force will be static friction and the contact point of the ball with the plane is at instantaneous rest
The point of contact is not slipping on the ground so we can say that the friction force work done would be zero.
So here in this case of pure rolling we can use the energy conservation
Part b)
By energy conservation principle we know that
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
so we will have
![\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega'^2 + mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%27%5E2%20%2B%20mgh)
here in pure rolling we know that
![v = R\omega](https://tex.z-dn.net/?f=v%20%3D%20R%5Comega)
now from above equation we have
![\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv'^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v'}{R})^2 + mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B2%7D%7B5%7DmR%5E2%29%28%5Cfrac%7Bv%7D%7BR%7D%29%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7B2%7D%7B5%7DmR%5E2%29%28%5Cfrac%7Bv%27%7D%7BR%7D%29%5E2%20%2B%20mgh)
now we have
![\frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv'^2(1 + \frac{2}{5}) + mgh](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%281%20%2B%20%5Cfrac%7B2%7D%7B5%7D%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%281%20%2B%20%5Cfrac%7B2%7D%7B5%7D%29%20%2B%20mgh)
now plug in all values in it
![\frac{1}{2}m(6^2)(\frac{7}{5}) = \frac{1}{2}mv'^2(\frac{7}{5}) + m(9.81)(0.80)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm%286%5E2%29%28%5Cfrac%7B7%7D%7B5%7D%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%27%5E2%28%5Cfrac%7B7%7D%7B5%7D%29%20%2B%20m%289.81%29%280.80%29)
![25.2 = 0.7v^2 + 7.848](https://tex.z-dn.net/?f=25.2%20%3D%200.7v%5E2%20%2B%207.848)
![v = 4.98 m/s](https://tex.z-dn.net/?f=v%20%3D%204.98%20m%2Fs)
It’s says that the temperature in the room is 23C so the air temperature in the air would be 23C