Answer:
Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.
Explanation:
So, we are given the following set of infomation in the question given above;
=> "spherical Gaussian surface of radius R centered at the origin."
=> " A charge Q is placed inside the sphere."
So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?
The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;
REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.
Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.
Answer:
From 0 -4 seconds the acceleration is positive. (The graph is going upwards.)
From 6-10 seconds the acceleration is negative. (The graph is going downwards.)
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)
Answer:
Reducing Discrimination by Changing Social Norms
Reducing Prejudice through Intergroup Contact
Explanation:
