Answer:
The answer is Letter B The car travel at a constant veloc
<span>Answer: Answer is The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.</span>
Answer:

Explanation:
As we know,
=》Force = Mass × Acceleration
=》45 N = 0.75 × Acceleration
=》Acceleration = 45 ÷ 0.75
=》Acceleration = 60
hence, the Acceleration of the ball would be. 60 meters per second square

Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
A. 
The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

where we have:
is the orbital speed
r is the orbital radius
is the orbital period
Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

B. 
The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

where
m is the mass of the clumps of matter
G is the gravitational constant
M is the mass of the black hole
Solving the formula for M, we find the mass of the black hole:

and considering the value of the solar mass

the mass of the black hole as a multiple of our sun's mass is

C. 
The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

where M is the mass of the black hole and c is the speed of light.
Substituting numbers into the formula, we find
