Given :
You fire a paintball from a paintball gun straight up into the air at 25.0 m/s.
To Find :
What is the ball velocity at the higher point that the ball reaches.
Solution :
We know, when ball is moving upward acceleration due to gravity is applied downward.
So, the ball will decelerate and when ball reaches maximum height its velocity will become zero.
Therefore, ball velocity is 0 m/s at maximum height.
Hence, this is the required solution.
Answer:
288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.
Explanation:
If all other parameters are constant,
Electrostatic Force of attraction ∝ (1/r²)
F = (k/r²) = 72.0
If r₁ = r/2, what happens to F₁
F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units
Answer:
Tension in the supporting cable is = 4,866 N ≅4.9 KN
Explanation:
First of all, we need to understand that tension is a force, so the motion law
F = Ma applies perfectly.
From Newtons third law of motion, action and reaction are equal and opposite. This means that the force experienced by the elevator, is equal to the tension experienced by the spring.
Parameters given:
Mass of load = 1650 kg
Acceleration of load = ?
The acceleration of the load can be obtained by diving the change in velocity by the time taken. But we need to know the time taken for the motion to 41 m.
Time taken = distance covered / velocity
=
= 3.73 seconds
∴Acceleration = ( initial velocity - final velocity )/ time taken
Note: Final velocity is = 0 since the body came to a rest.
Acceleration =
= 2.95m/
Force acting on the cable = mass of elevator × acceleration of elevator
= 1650 × 2.95 = 4869.5 kg ≅ 4.9 KN
Answer:
volume
Explanation:
it depends on the volume of the items
hope this helps
Answer:
Explanation:
The speed of the water in the large section of the pipe is not stated
so i will assume 36m/s
(if its not the said speed, input the figure of your speed and you get it right)
Continuity equation is applicable for ideal, incompressible liquids
Q the flux of water that is Av with A the cross section area and v the velocity,
so,


the diameter decreases 86% so


Thus, speed in smaller section is 48.6 m/s