Answer:
The linear momentum is zero
Explanation:
Because
When a rigid body is rotating about a fixed axis passing through point O, the body’s linear momentum given as L = mvG
But VG= 0 so
Linear momentum is zero
Answer:
![\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cfrac%7Bq%20ln%28%5Cfrac%7Bb%7D%7Ba%7D%29%7D%7B2%5Cpi%20%5Cepsilon_0%20L%7D)
Explanation:
As we know that the charge per unit length of the long cylinder is given as
![\lambda = \frac{q}{L}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bq%7D%7BL%7D)
here we know that the electric field between two cylinders is given by
![E = \frac{2k\lambda}{r}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B2k%5Clambda%7D%7Br%7D)
now we know that electric potential and electric field is related to each other as
![\Delta V = - \int E.dr](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20-%20%5Cint%20E.dr)
![\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20-%5Cint_a%5Eb%20%28%5Cfrac%7B2k%5Clambda%7D%7Br%7D%29dr%20)
![\Delta V = -2k \lambda ln(\frac{b}{a})](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20-2k%20%5Clambda%20ln%28%5Cfrac%7Bb%7D%7Ba%7D%29)
![\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cfrac%7B%5Clambda%20ln%28%5Cfrac%7Bb%7D%7Ba%7D%29%7D%7B2%5Cpi%20%5Cepsilon_0%7D)
![\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cfrac%7Bq%20ln%28%5Cfrac%7Bb%7D%7Ba%7D%29%7D%7B2%5Cpi%20%5Cepsilon_0%20L%7D)
Answer:
![\boxed{\sf 20 \$ = 80q}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%2020%20%5C%24%20%3D%2080q%7D%20)
Given:
1$ = 4q
To Find:
How many quarters are in 20$
Explanation:
To find out how many quarters are in 20$ we need to multiple 4 × 20.
![\sf 1\$ = 4q](https://tex.z-dn.net/?f=%20%5Csf%201%5C%24%20%20%3D%204q)
![\sf 20\$ = 4 \times 20q](https://tex.z-dn.net/?f=%20%5Csf%20%2020%5C%24%20%20%3D%204%20%5Ctimes%2020q)
![\sf = 80q](https://tex.z-dn.net/?f=%20%5Csf%20%3D%2080q)
Answer:
Explanation:
Given that,
Number of extra electrons, n = 21749
We need to find the net charge on the metal ball. Let Q is the net charge.
We know that the charge on an electron is
To find the net charge if there are n number of extra electrons is :
Q = n × q
So, the net charge on the metal ball is
. Hence, this is the required solution.
Answer:
Explanation:
The standard equation of the sinusoidal wave in one dimension is given by
![y = A Sin\left ( \frac{2\pi }{\lambda }\left ( vt-x \right )+\phi \right )](https://tex.z-dn.net/?f=y%20%3D%20A%20Sin%5Cleft%20%28%20%5Cfrac%7B2%5Cpi%20%7D%7B%5Clambda%20%7D%5Cleft%20%28%20vt-x%20%5Cright%20%29%2B%5Cphi%20%20%5Cright%20%29)
Here, A be the amplitude of the wave
λ be the wavelength of the wave
v be the velocity of the wave
Φ be the phase angle
x be the position of the wave
t be the time
this wave is travelling along positive direction of X axis
The frequency of wave is f which relates with velocity and wavelength as given below
v = f x λ
The relation between the time period and the frequency is
f = 1 / T.