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3 years ago

Which statement is true about an object moving in a circular motion due to centripetal force, F, when the mass is doubled?

Physics

1 answer:

marshall27 [118]3 years ago

4 0

Ans) A) Centripetal force will be doubled.

See centripetal force F = mv^2/r

That means centripetal force is directly proportional to the mass of the particle

So, if we double the mass, centripetal force will be increased by twofolds.

So, option A) is correct.

Now, looking at the other options,

B) says centripetal force is unaltered which is incorrect as centripetal force has been altered and increased twofold.

Option C) and D) reduces centripetal force which are also not possible here.

So, only Option A) is correct

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The energy from the sun that reaches the corn is about two billionths.

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An 4-kg ball experiences a force and accelerates at a rate of 1.5 m/s.

ANTONII [103]

Answer:

6.0 N

Explanation:

The strength of a force is expressed as the magnitude of the force in Newton.

The formula to apply here is :

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3 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

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3 years ago

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Answer:

TRUE

Explanation:

The gravitational force between two objects becomes weaker if the two objects are moved apart and stronger if they are brought closer together; that is, the force depends on the distance between the objects

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Elena-2011 [213]

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Explanation:

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3 years ago