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3 years ago

Wind tends to move from pressure areas. low, high high, high low, low high, low

Physics

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

high, low

Explanation:

Energy always flows from a higher level to a lower level.
It is analogous to the waterfall where waterfalls from a higher level to a lower level.
So in the case of the pressure of the gas, when there are any numbers of molecules in a given volume of space. The gas is said to be at high pressure.
When there are fewer molecules in the given volume. The gas is said to be at lower pressure.
Due to a large number of atoms, the high-pressure gas exerts more force on the container than the force exerted by the low-pressure gas.
If a hose is connected between these two containers, gas rushes from high pressure to the low pressure. Since the force exerted by the high-pressure gas is greater than that of low-pressure gas.

So, the wind tends to move from high-pressure areas to low pressure.

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A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

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A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with th

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