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3 years ago

The air in a cool region underneath cloud cover will have _____ a region with no cloud cover.

Physics

2 answers:

Slav-nsk [51]3 years ago

7 0

<u>Answer:</u>

The air is a <u>homogeneous mixture of gases</u> that make up the Earth's atmosphere, which remain around the planet by the attraction force of gravity.

Its properties changes with the altitude, the pressure, the temperature and the density. For example, there is a relation that is <u>inversely proportional</u> between the temperature and the air density.

<h2>When the temperature decreases the density increases. </h2>

Therefore, the air in a cool region of the atmosphere will be denser than the air in a hot region of the atmosphere.

In other words:

<h2>The air in a cool region underneath cloud cover will have <em><u>a higher density than</u></em> a region with no cloud cover </h2><h2> </h2>

Aleksandr-060686 [28]3 years ago

5 0

The air in a cool region underneath cloud cover will have a higher density than a region with no cloud cover. A region with no cloud cover has warmer air than the surrounding air, it becomes less dense and begins to rise, drawing more air in underneath.

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Can you guys please help me with this science quiz

Xelga [282]

Answer:

1 is 90, 2 is 200 and 3 is 5

Explanation:

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

Why can friction make observing Newton's first law of motion difficult?

Kamila [148]

The law says things travel in a straight line at constant speed unless acted upon by a force. But friction is a force but it can't be seen easily other than its effect, which is to bring the object to rest in seeming violation of the 1st law.

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3 years ago

A motor produces less mechanical energy than the energy it uses because the motor?

In-s [12.5K]

<span>A motor produces less mechanical energy than the energy it uses because the motor looses some energy to heat.</span>

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3 years ago

Acceleration is the magnitude of average velocity.

lina2011 [118]

Answer:

false

Explanation:

just did the question on apex, true was wrong

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3 years ago