Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
![\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3D-k%5Cfrac%7Bq%7D%7Br%5E2%7Dcos%5Ctheta%5C%20%5Chat%7Bi%7D%2Bk%5Cfrac%7Bq%7D%7Br%5E2%7Dsin%5Ctheta%5C%20%5Chat%7Bj%7D%5C%5C%5C%5C%5Cvec%7BE%7D%3Dk%5Cfrac%7Bq%5E2%7D%7Br%7D%5B-cos%5Ctheta%5C%20%5Chat%7Bi%7D%2Bsin%5Ctheta%5C%20%5Chat%7Bj%7D%5D)
(1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
The distance r is:
You replace the values of all parameters in the equation (1):
![\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5Cfrac%7B4.28%2A10%5E%7B-12%7DC%7D%7B%2810.21%2A10%5E%7B-3%7Dm%29%7D%5B-cos%2815.84%5C%C2%B0%29%5Chat%7Bi%7D%2Bsin%2815.84%5C%C2%B0%29%5Chat%7Bj%7D%5D%5C%5C%5C%5C%5Cvec%7BE%7D%3D%28-3.61%5Chat%7Bi%7D%2B1.02%5Chat%7Bj%7D%29%5Cfrac%7BN%7D%7BC%7D%5C%5C%5C%5C%7C%5Cvec%7BE%7D%7C%3D%5Csqrt%7B%283.61%29%5E2%2B%281.02%29%5E2%7D%5Cfrac%7BN%7D%7BC%7D%3D3.75%5Cfrac%7BN%7D%7BC%7D)
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
Answer:a. Magnetic dipole moment is 0.3412Am²
b. Torque is zero(0)N.m
Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A
That is,
U = n*I*A
But Area A is given as pi*radius² since it is a circular coil
Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is
A = 3.142*(0.05)² =7.86*EXP {-3} m²
Current I is 2 A
Number of turns is 20
So magnetic dipole moment U is
U = 20*2*7.86*EXP {-3}=0.3142A.m²
b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U
Torque = B x U =B*U*Sine(theta)
But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m
Answer:
The magnitude of the applied torque is 
(e) is correct option.
Explanation:
Given that,
Mass of object = 3 kg
Radius of gyration = 0.2 m
Angular acceleration = 0.5 rad/s²
We need to calculate the applied torque
Using formula of torque
Here, I = mk²
Put the value into the formula
Hence, The magnitude of the applied torque is
If the temperature is high there is less water because it evaporates if it is cloudy it is more because it doesn't evaporate
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
= initial volume = 5.68 L
= final volume = 2.35 L
Putting values in above equation, we get:
To convert this into joules, we use the conversion factor:
So, 
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
