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Fofino [41]
3 years ago
7

Is it possible to do work on an object without changing the kinetic energy of the object? Now Why?

Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

(a) Yes, it is possible by raising the object to a greater height without acceleration.

Explanation:

The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in  kinetic energy requires a change in velocity.

If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.

If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).

Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised  to a greater height without acceleration.

Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".

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A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th
Aleksandr [31]

Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

V = 200\,cm^{3}+800\,cm^{3}

V = 1000\,cm^{3}

The total volume of the air is 1000 cubic centimeters.

7 0
3 years ago
A force of 600 N is acting on a motorcycle that has a mass of 240 kg. What is the acceleration of the motorcycle?
Verdich [7]

Answer:

2.5m/s2

Explanation:

The following were obtained from the question:

F = 600N

M = 240 kg

a =?

Recall: F = Ma

a = F/M

a = 600/240

a = 2.5m/s2

Therefore, the acceleration of the motorcycle is 2.5m/s2

7 0
3 years ago
Read 2 more answers
Two hockey players have a total momentum of +200 kg x m/s before a collision (+ is to the right). After the collision, they move
Juliette [100K]

Total momentum after the collision: +200 kg m/s to the right

Explanation:

We can answer this question by using the law of conservation of momentum, which states that for an isolated system (=no external forces acting on the system), the total momentum is conserved.

Mathematically,

p_i=p_f

where

p_i is the total momentum before the collision

p_f is the total momentum after the collision

In this problem, the system consists of two hockey players. Before the collision, their total momentum is

p_i = +200 kg m/s (to the right)

Therefore, according to the law of conservation of momentum, their total momentum after the collision must be the same:

p_f = +200 kg m/s

And given that the sign is +, the direction is still the same, therefore to the right.

Learn more about momentum:

brainly.com/question/7973509

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#LearnwithBrainly

8 0
3 years ago
A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio
goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

3 0
3 years ago
Imagine that you’re observing a collision. Which action would allow you to determine whether the collision is inelastic?
Thepotemich [5.8K]

The answer is B;) in plato

3 0
3 years ago
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