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fiasKO [112]
1 year ago
5

How does the increasing mass effect the force of an object in motion?​

Physics
1 answer:
irina [24]1 year ago
7 0

Answer:

<u>According </u><u>to </u><u>second </u><u>law </u><u>of </u><u>motion</u><u>,</u><u>t</u><u>he acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.</u>

<em>So </em><em>simply</em><em>,</em><em> </em><em>it </em><em>can </em><em>be </em><em>affected </em><em>due </em><em>to </em><em>increasing </em><em>force </em><em>as </em><em>there </em><em>is </em><em>close </em><em>relationship </em><em>between </em><em>momentum.</em>

Explanation:

<em>The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.</em>

<em>I </em><em>hope </em><em>it </em><em>was </em><em>helpful </em><em>for </em><em>you </em><em>:</em><em>)</em>

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A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
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Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

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Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

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4 0
3 years ago
A force of 200 N stretches a spring 30 cm. What is the spring constant of the spring? How far would this spring stretch with a f
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Hooke's Law

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Δx = 30 cm = 0.3 m

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the spring stretch for 100 N:

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An ice skater has a moment of inertia of 5.0 kg · m2when her arms are outstretched, and at this time she is spinning at 3.0 rev/
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Answer:

Explanation:

Given

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Conserving Angular momentum

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N_2=\frac{\omega _2}{2\pi }

N_2=\frac{15\pi }{2\pi}=7.5 rev/s

8 0
3 years ago
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