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:
T1 = 130N, T2 = 370N
Explanation:
In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.
1. forces:
Let tension in rope 1 be T1 and in rope 2 be T2:
ma = T1 + T2 - 100N - 400N = 0
(1) T1 + T2 = 500N
2. torque around the center point of the beam:
τ = r x F = 5*T1 + 3*400N - 5*T2 = 0
(2) T1 - T2 = -240N
Solving both equations:
T1 = 130N
T2 = 370N
Answer:
Explanation:
There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.
momentum of first piece = .320 kg x 2 m/s
= 0.64 kg m/s along x -axis.
momentum of second piece = .355 kg x 1.5 m/s
= 0.5325 kg m/s along y- axis .
Let the velocity of third piece be v and it is making angle of θ with x -axis .
Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ
vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ
For making total momentum in the plane zero
.1 v cosθ = 0.64 kg m/s
.1 v sinθ = 0.5325 kg m/s
Dividing
Tanθ = .5325 / .64 = .83
θ = 40⁰.
The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.
they are the only ones with a cell wall.
If ball remains in air for total time T = 0.85 s
this is also known as time of flight
In order to find the time of flight we can use kinematics
so for complete motion its displacement in y direction will be zero
now we know that net velocity of the ball is 8 m/s
while is y direction component we got is vy = 4.165 m/s
now by component method we can say
so it is projected at an angle of 31.4 degree above horizontal
