It'll last 20 hours. If it travels 110 miles in one hours, 110 times 20 equals 2,200.
Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor, ![Q=121\ nC=121\times 10^{-9}\ C](https://tex.z-dn.net/?f=Q%3D121%5C%20nC%3D121%5Ctimes%2010%5E%7B-9%7D%5C%20C)
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
![Q=ne](https://tex.z-dn.net/?f=Q%3Dne)
e is the charge on electron
![n=7.5625\times 10^{11}](https://tex.z-dn.net/?f=n%3D7.5625%5Ctimes%2010%5E%7B11%7D)
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
![q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.](https://tex.z-dn.net/?f=q_1%3D-4.10%5C%20nC%3D-4.10%5Ctimes%2010%5E%7B-9%7D%5C%20C.)
- Charge on the second charged particle,
![q_2=3.80\ nC=3.80\times 10^{-9}\ C.](https://tex.z-dn.net/?f=q_2%3D3.80%5C%20nC%3D3.80%5Ctimes%2010%5E%7B-9%7D%5C%20C.)
- Position of the first charge =
![(x_1=0.00\ m,\ y_1=0.600\ m).](https://tex.z-dn.net/?f=%28x_1%3D0.00%5C%20m%2C%5C%20y_1%3D0.600%5C%20m%29.)
- Position of the second charge =
![(x_2=1.50\ m,\ y_2=0.650\ m).](https://tex.z-dn.net/?f=%28x_2%3D1.50%5C%20m%2C%5C%20y_2%3D0.650%5C%20m%29.)
The electric field at a point due to a charge
at a point
distance away is given by
![\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%5Cdfrac%7Bkq%7D%7B%7C%5Cvec%20r%7C%5E2%7D%5C%20%5Chat%20r.)
where,
= Coulomb's constant, having value ![\rm 8.99\times 10^9\ Nm^2/C^2.](https://tex.z-dn.net/?f=%5Crm%208.99%5Ctimes%2010%5E9%5C%20Nm%5E2%2FC%5E2.)
= position vector of the point where the electric field is to be found with respect to the position of the charge
.
= unit vector along
.
The electric field at the origin due to first charge is given by
![\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.](https://tex.z-dn.net/?f=%5Cvec%20E_1%20%3D%20%5Cdfrac%7Bkq_1%7D%7B%7C%5Cvec%20r_1%7C%5E2%7D%5C%20%5Chat%20r_1.)
is the position vector of the origin with respect to the position of the first charge.
Assuming,
are the units vectors along x and y axes respectively.
![\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.](https://tex.z-dn.net/?f=%5Cvec%20r_1%3D%280-x_1%29%5Chat%20i%2B%280-y_1%29%5Chat%20j%5C%5C%3D%280-0%29%5Chat%20i%2B%280-0.6%29%5Chat%20j%5C%5C%3D-0.6%5Chat%20j.%5C%5C%5C%5C%7C%5Cvec%20r_1%7C%20%3D%200.6%5C%20m.%5C%5C%5Chat%20r_1%3D%5Cdfrac%7B%5Cvec%20r_1%7D%7B%7C%5Cvec%20r_1%7C%7D%3D%5Cdfrac%7B0.6%5C%20%5Chat%20j%7D%7B0.6%7D%3D-%5Chat%20j.)
Using these values,
![\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.](https://tex.z-dn.net/?f=%5Cvec%20E_1%20%3D%20%5Cdfrac%7B%288.99%5Ctimes%2010%5E9%29%5Ctimes%20%28-4.10%5Ctimes%2010%5E%7B-9%7D%29%7D%7B%280.6%29%5E2%7D%5C%20%28-%5Chat%20j%29%3D1.025%5Ctimes%2010%5E2%5C%20N%2FC%5C%20%5Chat%20j.)
The electric field at the origin due to the second charge is given by
![\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.](https://tex.z-dn.net/?f=%5Cvec%20E_2%20%3D%20%5Cdfrac%7Bkq_2%7D%7B%7C%5Cvec%20r_2%7C%5E2%7D%5C%20%5Chat%20r_2.)
is the position vector of the origin with respect to the position of the second charge.
![\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.](https://tex.z-dn.net/?f=%5Cvec%20r_2%3D%280-x_2%29%5Chat%20i%2B%280-y_2%29%5Chat%20j%5C%5C%3D%280-1.50%29%5Chat%20i%2B%280-0.650%29%5Chat%20j%5C%5C%3D-1.5%5Chat%20i-0.65%5Chat%20j.%5C%5C%5C%5C%7C%5Cvec%20r_2%7C%20%3D%20%5Csqrt%7B%28-1.5%29%5E2%2B%28-0.65%29%5E2%7D%3D1.635%5C%20m.%5C%5C%5Chat%20r_2%3D%5Cdfrac%7B%5Cvec%20r_2%7D%7B%7C%5Cvec%20r_2%7C%7D%3D%5Cdfrac%7B-1.5%5Chat%20i-0.65%5Chat%20j%7D%7B1.634%7D%3D-0.918%5C%20%5Chat%20i-0.398%5Chat%20j.)
Using these values,
![\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.](https://tex.z-dn.net/?f=%5Cvec%20E_2%3D%20%5Cdfrac%7B%288.99%5Ctimes%2010%5E9%29%5Ctimes%20%283.80%5Ctimes%2010%5E%7B-9%7D%29%7D%7B%281.635%29%5E2%7D%28-0.918%5C%20%5Chat%20i-0.398%5Chat%20j%29%20%3D-11.74%5C%20%5Chat%20i-5.09%5C%20%5Chat%20j%5C%20%20N%2FC.)
The net electric field at the origin due to both the charges is given by
![\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%5Cvec%20E_1%2B%5Cvec%20E_2%5C%5C%3D%28102.5%5C%20%5Chat%20j%29%2B%28-11.74%5C%20%5Chat%20i-5.09%5C%20%5Chat%20j%29%5C%5C%3D-11.74%5C%20%5Chat%20i%2B%28102.5-5.09%29%5Chat%20j%5C%5C%3D%28-11.74%5C%20%5Chat%20i%2B97.41%5C%20%5Chat%20j%29%5C%20N%2FC.)
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow
![m1U1+ m2U2= m1V1+m2V2](https://tex.z-dn.net/?f=m1U1%2B%20m2U2%3D%20m1V1%2Bm2V2)
Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow
![m1U1+ m2U2= V(m1+m2)](https://tex.z-dn.net/?f=m1U1%2B%20m2U2%3D%20V%28m1%2Bm2%29)
Answer: ![1.298\times 10^{-5}\ N/m](https://tex.z-dn.net/?f=1.298%5Ctimes%2010%5E%7B-5%7D%5C%20N%2Fm)
Explanation:
Given
Current in the first wire ![I_1=2.2\ A](https://tex.z-dn.net/?f=I_1%3D2.2%5C%20A)
Current in the second wire ![I_2=5.9\ A](https://tex.z-dn.net/?f=I_2%3D5.9%5C%20A)
wires are
apart
Force per unit length between the current-carrying wires is
![\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cdfrac%7BF%7D%7Bl%7D%3D%5Cdfrac%7B%5Cmu_oI_1I_2%7D%7B2%5Cpi%20r%7D)
Force exerted by the wires is the same
Put the values
![\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7BF%7D%7Bl%7D%3Df%3D%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%202.2%5Ctimes%205.9%7D%7B2%5Cpi%20%5Ctimes%200.2%7D%3D1.298%5Ctimes%2010%5E%7B-5%7D%5C%20N%2Fm)
This force will be repulsive in nature as the current is flowing opposite