Answer: Did you get the answers? Im struggling homie.
Explanation:
Knowing the direction of a force is important because it helps someone know the motion of the object. if you use a free body diagram, then it becomes easy to see all the forces being applied to an object. if there is more force going one way, the object is accelerating in that direction. if all the forces cancel each other out, then the object is at a constant speed or is at rest.
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:
(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:
Current for 12 ohms = 2.5
Explanation:
Ohm's Law:
Voltage = IR
Current = V/R
Resistance = V/I
Therefore, current = 30V/12 Ohms
30/12 = 2.5
So current for 12 ohms = 2.5
I hope this helps :>
Answer:
Rockets provide a wonderful example of Momentum Conservation. As momentum in one direction is given to the rocket's exhaust gases, momentum in the other direction is given to the rocket itself.
Explanation:
First, think of two masses connected by a lightweight (massless!) compressed spring. When the two spring apart, conservation of momentum tells us the Center of Mass remains where it was (or moving as it was).
PTot,i = p1i + p2i = 0 + 0 = 0
PTot,f = p1f + p2f = PTot,i = 0
p1f + p2f = - m1 v1f + m2 v2f = 0