Answer:
Explanation:
Given
Initial velocity u = 200m/s
Final velocity = 4m/s
Distance S = 4000m
Required
Acceleration
Substitute the given parameters into the formula
v² = u²+2as
4² = 200²+2a(4000)
16 = 40000+8000a
8000a = 16-40000
8000a = -39,984
a = - 39,984/8000
a = -4.998m/s²
Hence the acceleration is -4.998m/s²
Explanation:
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

At t = 1 s,
Current,

So, the current at t = 1 s is 3 A.
For lowest current,

Hence, this is the required solution.
Answer:
The maximum potential difference is 186.02 x 10¹⁵ V
Explanation:
formula for calculating maximum potential difference

where;
Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²
k is the dielectric constant = 2.3
b is the outer radius of the conductor = 3 mm
a is the inner radius of the conductor = 0.8 mm
λ is the linear charge density = 18 x 10⁶ V/m
Substitute in these values in the above equation;

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V
Choice-'a' is a slippery, misleading, ambiguous statement,
but it's less wrong than any of the other choices on this list.
Answer:
The object's velocity would increase due to the change in force.
Explanation: