What happens to the temperature as altitude increases in the exosphere? Does it increase or decrease the higher it goes?

When doing yoga stretches, you are working on the cardiovascular endurance component of physical fitness.

Tomtit [17]

I believe it is true :D

6 0

2 years ago

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A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.

Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is 1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

#SPJ1

6 0

2 years ago

What countries could benefit most from solar energy

Alecsey [184]

Countries longer day for example in the Middle East could benefit more but Alaska has more night like longer night which is why it would hard to benefit from the solar panels there.

Hoped this helped

5 0

2 years ago

Reduce the work output

andriy [413]

Maybe The 3rd One Not Sure!

6 0

3 years ago

Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm

Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

$I_1\omega_1 = I_2\omega_2$

now here we have

$I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2$

$I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2$

$I_1 = 256$

Now when children come to the position of half radius

then we will have

$I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2$

$I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2$

$I_2 = 160$

now from above equation we have

$256 (20.0 rpm) = 160(\omega_2)$

$\omega_2 = 32 rpm$

8 0

2 years ago