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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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The coefficient of static friction between the puck and the surface.

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2 years ago
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Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the
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Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, \theta=135^{\circ}

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

W=Fd\ cos\theta

W=50\times 9\times \ cos(135)

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

7 0
3 years ago
Two concrete spans of a 250-m-long bridge are placed end to end so that no room is allowed for expansion. If a temperature incre
scoundrel [369]

Answer:

  y = 2.74 m

Explanation:

The linear thermal expansion processes are described by the expression

         ΔL = α L ΔT

Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc

If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length

        ΔL = 12 10⁻⁶ 125 20

        ΔL = 3.0 10⁻² m

Let's use trigonometry to find the height

The hypotenuse     Lf = 125 + 0.03 = 125.03 m

Adjacent leg           L₀ = 125 m

       cos θ = L₀ / Lf

       θ = cos⁻¹ (L₀ / Lf)

       θ = cos⁻¹ (125 / 125.03)

       θ = 1,255º

We calculate the height

       tan 1,255 = y / x

       y = x tan 1,255

       y = 125 tan 1,255

       y = 2.74 m

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3 years ago
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Which of the following is not an example of a physical property?
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