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3 years ago

Which of the following best describes wind?

Physics

1 answer:

Afina-wow [57]3 years ago

5 0

Answer:

The answer is D the rising of warm air pushing down cool air.

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Which phenomenon would cease to exist in the absence of this axial tilt?

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If the Earth didn't tilt then we wouldn't have seasons.

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3 years ago

Which action might lead scientists to develop new explanations about the

VikaD [51]

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Explanation:

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3 years ago

Over thousands of years, when sediment is squeezed together and the pore space between the grains is reduced, this is the proces

Aneli [31]

This is the process of compaction.

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3 years ago

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

6 0

3 years ago

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5

Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

$v = \sqrt{\frac{F_T}{\mu}}$

where $F_T = 44N$ is the tension on the rope and $\mu = m/L = 0.5 / 2.6 = 0.1923 kg/m$ is the density of the rope per unit length

$v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s$

6 0

3 years ago