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1 year ago

Air that is not saturated will cool or heat at a rate of __________________ as it rises or descends, respectively

1 answer:

7 0

Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively

The rate of the change in temperature which is observed while moving upward in the earth's atmosphere with elevation. It can be positive, negative and zero when the temperature decreases, increases or is constant with the elevation respectively.

For the atmosphere, the drop in temperature of rising air that is unsaturated air is about 10 degrees C/1000 meters (5.5 °F per 1000 feet) altitude.

That means if a there occur a rise of 1000 m , then the temperature of that thing will decrease to 10 degrees. Every 10°C of temperature from the given temperature will decrease at every rise of 1000 m .Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively

To learn more about unsaturated air here

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How much gravitational potential energy does a 1.0 kg hammer have on a shelf 1.5 m above the ground?

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Answer:

14.7 J

Explanation:

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2 years ago

What will be the ME of a machine that produces a 240 N work with a 300

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Answer:

Efficiency = 80%

Explanation:

Given the following data;

Work output = 240 N

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$Efficiency = \frac {Out-put \; work}{In-put \; work} * 100$

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$Efficiency = \frac {240}{300} * 100$

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2 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

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Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$