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Anastaziya [24]

2 years ago

14

A ball has a kinetic energy of 106 J. If the ball is moving at a velocity of 2.01 m/s, what is the mass of the ball, in kilogram

s

1 answer:

dem82 [27]2 years ago

4 0

Answer:

Explanation:

KE= ½mv²

m = 2KE/v²

m = 2(106)/2.01²

m = 52.47394...

m = 52.5 kg

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Answer:

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Explanation:

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3 years ago

brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o

beks73 [17]

Maximum speed of the child as she swings is 2.23 m/s.

<h3>Step by Step Calculation:</h3>

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k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

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1 year ago

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

elena55 [62]

Answer:

The first part can be solved via conservation of energy.

$mgh = mg2R + K\\K = mg(h-2R)$

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

$F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}$

where $N = 0$ because we are looking for the case where the car loses contact.

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Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

$mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}$

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0

2 years ago

This question is related to inertia:

luda_lava [24]

The way I do it is suddenly, in the same sort of way that magicians try to pull a table cloth off a table when there's things on the table cloth.The sudden approach acts as an impulse of force and starts to accelerate the roll. But, the piece (assuming it has perforations) is off the roll before the roll can move, due to inertia. Then the roll will acclerate, move, slow down and stop. However, in accelerating, the roll will unravel. The bigger the impulse the more it will unravel.+++++++++++++++++++++++++++++++++++++++If on the other hand, the piece of paper is held firmly, and the roll is pulled, then the impulse is presumably given to the paper and the hand whose inertia is a lot more than that of the roll. So, I think I'd actually go for choice c)+++++++++++++++++++++++++++++++++++++This assumes that the roll is free to rotate.I think that a similar idea is behind the design and use of a "ballistic galvanometer". The charge is passed through the galvanometer quickly, as a current pulse. Then the needle starts to deflect, and the deflection is arranged to depend on the total charge that has passed through in the time of the current pulse.

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2 years ago