Answer:
When two spheres, each with charge Q, are positioned a distance Rapart, they are attracted to ... doubled, the electric-force between the two spheres
The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
CALCULATE INITIAL VELOCITY:
The initial velocity of the car can be calculated by using one of the equation of motion as follows:
V = u + at
Where;
- V = final velocity (m/s)
- u = initial velocity (m/s)
- a = acceleration due to gravity (m/s²)
- t = time (s)
According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:
u = v - at
u = 27 - 3(5)
u = 27 - 15
u = 12m/s.
Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
Learn more about motion at: brainly.com/question/974124
Answer:
We first to know that if the wheel rotates from rest means that at t=0 the velocity and the angle rotated is 0.
Then, we know:

Integrating 2 times, we have:

For the first 27.9 s, we have:
w = 37.107 rad/s
angle = 517.6426 rad
For the next seconds, according to the text, the angular velocity is constant so
w = 37.107 rad/s and hence, integrating:

Then, the time remaining is:
53.5 - 27.9 = 25.6
So for the next 25.6 seconds we have:

Finally, we add the 2 angles and we have as a result:

Answer:
Explanation:
Speed given = 125 m /min
125 /60 m /s
In 450 second it will travel
= 450 x 125 / 60
=937.5 m.
As the distance is covered in less than 450 seconds , The distance must be less than 937.5 m
In 400 seconds , it will travel
= 400 x 125 / 60
833.33 m
Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .
Hence the distance covered is more than .833 m but less than 937.5
In either case these distance are more than .8 km .
Answer:
a) 4.45 m/s
b) 0.9 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²

a) The vertical speed when the player leaves the ground is 4.45 m/s

Time taken to reach the maximum height is 0.45 seconds

Time taken to reach the ground from the maximum height is 0.45 seconds
b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds