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Stolb23 [73]
3 years ago
5

How much momentum will a dumb-bell of mass 10 kg transfer

Physics
1 answer:
frosja888 [35]3 years ago
6 0

We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:

Vf² = Vi² + 2ad

Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.

Given values:

Vi = 0m/s (dumbbell starts falling from rest)

a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)

d = 80×10⁻²m

Plug in the values and solve for Vf:

Vf² = 2(10)(80×10⁻²)

Vf = ±4m/s

Reject the negative root.

Vf = 4m/s

The momentum of the dumbbell is given by:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

m = 10kg

v = 4m/s (from previous calculation)

Plug in the values and solve for p:

p = 10(4)

p = 40kg×m/s

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An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
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Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

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b)

frequency= f = 1/ T

f = 1/ 0.372

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c).

Angular frequency= w= 2πf

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d)

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As w=\sqrt{k/m}

⇒w²= k/m

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⇒k= 0.628 * 16.9² N/m

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e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

⇒ v= 8.78 m/s

f)

Maximum force will be exerted on the block when it is at maximum distance.

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Answer:

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The  sphere have some mass at the center but most of its mass is closer to its radius and thus have more inertia than flat Disk.

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Answer:

Please find the answer in the explanation

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