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Olegator [25]
3 years ago
11

The quarterback throws a football at 25 m/s at a certain angle above the horizontal. If it took the ball 2.5 s to reach the top

of its path, how long was it in the air?
Physics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

The ball was in the air for 5.0 seconds

Explanation:

When the ball is thrown above at a certain angle from the horizontal it follows a projectile motion.

The ball has a specific vertical and horizontal velocity.

The time take for the ball to reach at the top of its path is also the time taken by the vertical velocity vector to become zero. This is the first half of the projectile motion (ball is going up).

For the second half, the ball comes down and it accelerates with same gravitational acceleration from which it was decelerating in the first half. Hence the time to come down will be the same 2.5 seconds.

So the total time the ball was in the air is twice of 2.5, i.e,

5.0 seconds

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USPshnik [31]

Answer:

The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.

Explanation:

The detailed solution can be found in the attachment below.

Thank you for reading and I hope this is helpful to you.

6 0
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A bird accelerating from rest at a constant rate,experiences a displacement of 28 m in 11s.What is its acceleration
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Use the formula below for this question:

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The ideal gas law includes all of the following properties, except the
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How should you draw the field lines for Earth's magnetic field?
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Read 2 more answers
n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad
Oksi-84 [34.3K]

Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

F1/A1 = F2/A2

From our question,

The radius of the smaller piston is 5 cm = 0.05 m

The radius of the larger piston is 15 cm = 0.15 m

The force of the larger piston is 13300 N

The force of the smaller piston is unknown = F

A1 = πr² = 3.142 * 0.05² = 0.007855 m²

A2 = πr² = 3.142 * 0.15² = 0.070695 m²

F1/0.007855 = 13300/0.070695

F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

5 0
3 years ago
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