Answer:
The speed of the ball was, v = 3 m/s
Explanation:
Given data,
The time period of the ball, t = 8 s
The distance the ball rolled, d = 24 m
The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,
v = d / t m/s
Substituting the given values in the above equation,
v = 24 / 8
= 3 m/s
Hence, the speed of the ball was, v = 3 m/s
The gravitational force between the two balls is 
Explanation:
The magnitude of the gravitational force between two objects is given by:
where
:
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between the objects
For the balls in this problem, we have


r = 0.74 m
Substituting into the equation, we find the gravitational force between the two balls:
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Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
Answer:

Explanation:
Given that,
Initially, the spaceship was at rest, u = 0
Final velocity of the spaceship, v = 11 m/s
Distance accelerated by the spaceship, d = 213 m
We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

So, the acceleration experienced by the occupants of the spaceship is
.