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Anni [7]
3 years ago
8

Change in v= 9.8 m/s2xt. The diagram shows a ball falling toward Earth in a vacuum.

Physics
2 answers:
Tresset [83]3 years ago
5 0

Answer:

Option A. 39.2 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 4 s

Final velocity (v) =?

v = u + gt

Since the initial velocity (u) is 0, the above equation becomes:

v = gt

Thus, inputting the value of g and t, we can obtain the value of v as shown below:

v = 9.8 × 4

v = 39.2 m/s

Therefore, the velocity of the ball at 4 s is 39.2 m/s.

aliya0001 [1]3 years ago
5 0

39.2 m/s

Explanation

a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s/s in The Physics Classroom Tutorial in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy.

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True

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Explanation:

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7 0
3 years ago
a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d
USPshnik [31]

Answer:

The time is t = 2.595 \  s

The speed is v = 25.43 \ m/s

Explanation:

From the question we are told that

    The height of the cliff is  h =  33 \  m

 Generally from kinematic equation we have that

       h  =  ut + \frac{1}{2} gt^2

before the jump the persons initial velocity is  u =  0 m/s

 So

        33   =  0 * t + \frac{1}{2} 9.8 * t^2

=>      t = 2.595 \  s

Generally from kinematic equation

     v= u + gt

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=>  v = 25.43 \ m/s

3 0
3 years ago
A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How
nalin [4]

Answer:

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

6 0
3 years ago
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