v = 400m/58s = 6.9m/s
v = 400m/60s = 6.7m/s
6.9m/s - 6.7m/s = .2m/s difference
60sec - 58 sec = 2 sec
v =Δx/t
Δx = vt
Δx = (.2m/s)(2sec)
Δx = .4m
therefore, the answer is .4m
Answer:
Q = ba⁴ * ε₀
Explanation:
From Gauss's Law, we know that
flux Φ = Q / ε₀
where ε₀ = 8.85e-12 C²/N·m²
and also,
Φ = EAcosθ
The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus
Φ = EAcos0
Φ = EA
E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then
E = ba²
Since A = a² for the cube face, we have
Q / ε₀ = E * A
Q / ε₀ = ba² * a²
so that
Q = ba⁴ * ε₀
Not 100% but i think it'll cause the earth to rotate slightly slower, its definitely not the last one though
Answer:
The inducerd emf is 1.08 V
Solution:
As per the question:
Altitude of the satellite, H = 400 km
Length of the antenna, l = 1.76 m
Magnetic field, B = 
Now,
When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

Here, velocity v is perpendicular to the rod
Thus
e = lvB (1)
For the orbital velocity of the satellite at an altitude, H:

where
G = Gravitational constant
= mass of earth
= radius of earth

Using this value value in eqn (1):

Answer:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;

where;
a is the angular acceleration and
r is the radius of the circular path

Determine time of the rotation;

Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration

The magnitude of total linear acceleration is given by;

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²