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IrinaK [193]
2 years ago
15

If the mass of the object were doubled (keeping a constant radius on the string), what would have to happen to the period in ord

er to maintain the same tension on the string
Physics
1 answer:
Mrrafil [7]2 years ago
7 0

Answer:

the period of the motion will increase by√2.

Explanation:

Given that the motion you are talking about is circular motion of a mass attached to the end of a string. I speculated that from the word usage in the question(e.g radius instead of length, tension, period). Given this is so we will have to recall the formula for the centripetral force Fc acting on the object which will be equal in magnitude to the tension in the string and will be given by,

F_{c} = mrw^{2}

if we want the above defined tension to remain constant when we double the mass and keep the radius of the string constant, the the w(angular frequency) must change which is related to the period by the below equation which will also change,

w = 2\pi /T

to find out by how much the period will change we see that from the first equation that if we double the mass making it 2m then the <em>w</em>² will have to decrease by 2 that is it will become <em>w</em>²/2, at the same time keeping r constant since it says that in the question. We now absorb the 2 inside the square and we get,

F_{c} = 2mr(w/\sqrt{2} )^{2} =2mr(2\pi /T\sqrt{2} )^{2}

we can clearly see that the new period has become,

T_{new} = \sqrt{2} T

where T is the old period. So the new period is √2 times the old period given by the equation above.

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FrozenT [24]

Answer:

As the launch force increase the launch velocity will

<em><u>Increase</u></em>

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<em><u>There is a direct relationship between force and acceleration.</u></em>

<em><u /></em>

Explanation:

<em>It is known all over the place that, there is a direct relationship between Force and acceleration of an object leading to an increase in force being directly proportional to the increase in the acceleration of the given object and vice versa.</em>

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3 years ago
On a calm day with no wind, you can run a 1500-m race at a velocity of 4.0 m/s. If you run the same race on a day when you have
lesantik [10]

Answer:

The time taken to finish the race is 750 s.

Explanation:

The velocity of the person on the day of wind is slowed down by 2.0 m/s. So the person's velocity on the day of wind is 4-2=2 m/s.

The relation between time, speed and distance is t=v/d

Given d=1500 m and calculated v= 2 m/s.

t=1500/2

t=750 s.

Learn more about distance formula.

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3 0
1 year ago
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How would the number 13,900 be written using scientific notation?
Oksanka [162]

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um d. but I am guessing this ans

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2 years ago
Calculate the pressure exerted on the ground by a boy of a mass 60 kg if he stands on one foot.the area of the sole of his shoe
ddd [48]

Answer:

40 Kpa

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8 0
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ollegr [7]

Answer:

Explanation:

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As the volume of both the objects is same, so the buoyant force acting on both the objects is same.

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2 years ago
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