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sladkih [1.3K]
3 years ago
15

A physics student throws a ball straight up. The student catches the ball in exactly the same place from which it was released.

The ball’s time of flight is T, and its maximum height above its release point is H. Neglect air resistance and assume up is the positive direction. Find the ball’s average velocity during the second half of its trip. (Hint: Your answer should only have the variables H and T in it.)
Physics
1 answer:
mr Goodwill [35]3 years ago
4 0

Answer:

The correct answer is H ÷ ¹/₂T

Explanation:

The formula for velocity is distance covered ÷ time.

Neglecting air resistance;

If the ball's time of overall time flight is T, the time it will take for the second half/return trip is ¹/₂T.

If the ball's maximum height above its released point is H, the height will also be the distance it covered for the second part of the trip since the student caught the ball in the exact same place the ball was thrown. Hence, the distance for the second half of the trip will be H.

Since velocity = distance/time

The average velocity during the second half of the trip will be = H ÷ ¹/₂T

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The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is  110 dB.

Here we are dealing with loudness which is the perception of the Intensity of the sound.

The formula  to refer to in order  to  find the value of the loudness of a sound is ,

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As we are provided with the current intensity which is  1 x 10⁻¹ Wm⁻². and the initial intensity which is  1 x 10⁻¹² Wm⁻².

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1 year ago
An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti
Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

  • There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
  • Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
  • So, we can write the following expression for Fg:

       F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)

b)

  • The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

       F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N  (2)

c)

  • According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

       a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2  (3)

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<span> The masses have no inertia about their own CM, and "the object" is the two masses. </span>
<span>1. Icm (at point A) = 2mr^2 
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