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podryga [215]
3 years ago
8

A glass falls from a table top and smashes on the floor 0.6 seconds later. How high is the table?

Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

1.8 m

Explanation:

Given: Glass falls from a table, smashes 0.6 seconds later

To find: How high a table is

Formula: Vv=gt, dv=1/2gt^2, t=2d/g

Solution: A table's <em>height</em> is measured from the top of the edge down to the floor. The tables are shown both have a height of 30 inches, which is common for many tables.

<u>Data</u>

  • t = 0.6s
  • g = 9.81 m/s²
  • d = ?

<u>Equation</u>

  • d = \frac{1}{2}gt²

<u>Math & Units</u>

  • d = 4.905 (0.6²)
  • d = 442.676

Hence the table is 1.8 m high

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A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10
bearhunter [10]

Answer:

605447.7066 kgm²/s

Explanation:

m_1 = Mass of sphere = 10000 kg

m_2 = Mass of rod = 10 kg

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l = Length of antenna = 3 m

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\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s

Angular momentum is given by

L=I\omega

Moment of inertia of the satellite is

I_s=\frac{2}{5}m_1r^2

Moment of antenna of the satellite is

I_a=\frac{1}{3}m_2l^2

The angular momentum of the system is

L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s

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5 0
3 years ago
Which statement is true? A) The work done to lift an object 6 meters is greater than the gravitational potential energy it gains
inna [77]

The work required to raise an object to a height is equal to the gravitational potential energy the object gains. <em>(C)</em>

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Answer: A cold front occurs when a cold air mass advances into a region occupied by a warm air mass. If the boundary between the cold and warm air masses doesn't move, it is called a stationary front.

Explanation: Two types of occluded front exist: the warm-type and the cold-type. They’re distinguished by the relative temperatures of the air mass ahead of the occlusion – in other words, the air mass ahead of the original warm front – and the air mass behind the cold front. If the air behind the cold front is colder than the air ahead of the occlusion, it shoves beneath that air (because it’s denser) to form a cold-type occluded front. If the air behind the cold front is warmer than the air ahead, it rides over it to form a warm-type occluded front – which appears to be the more common case. In either situation, the lighter warm air representing the air mass originally between the warm and cold fronts sits above the boundary between the two cooler air masses.

Hope this helps!!

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