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3 years ago

What is the average velocity of the object?

Physics

1 answer:

Andrej [43]3 years ago

3 0

Answer:the rate changes during the position of the object

Explanation:so there is no object that has the same rate but unless it is a specific one like a care but it changes during the position of the object

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What is the acceleration of a body moving with uniform velocity?

AfilCa [17]

Acceleration is the rate of change of velocity, a body moving with uniform velocity does not possess acceleration at all i.e. acceleration is zero

3 0

3 years ago

You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. Wha

Rasek [7]

Answer:

K/2

Explanation:

The law of conservation of mechanical energy states that the sum of the kinetic and potential energies is a constant at any point.

At maximum height, the glove has purely potential energy but at the bottom, it has purely kinetic energy.

The potential energy at the top = kinetic energy at the bottom. The potential energy is given by

$PE = mgh$

At half height, this potential energy is

$PE = \frac{1}{2}mgh$

At this height, PE + KE = Constant = KE at bottom or PE at maximum height.

$mgh = \frac{1}{2}mgh +KE$

$KE = \frac{1}{2}mgh = K/2$

5 0

3 years ago

Momentum is a vector quantity that depends

MissTica

On the change in potential energy

3 0

3 years ago

Which set of terms is in the proper order, from left to right, for labeling the

Rufina [12.5K]

A. Base , salt , water,acid

3 0

3 years ago

A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the

Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

v = u + at

0 = 19.09 - 9.81 t

t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

S = ut + 0.5 at²

H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

Maximum height of the ball = 18.57 m

6 0

3 years ago