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Serjik [45]
3 years ago
9

Two clouds collide and form another, more massive cloud. One cloud is stationary, while the other is traveling at 1 m/s. After t

he collision, the new, combined cloud travels with a velocity of 0.25 m/s. What is the ratio of the masses of the two original clouds
Physics
1 answer:
sweet [91]3 years ago
6 0

Answer: 3

Explanation:

Given

One cloud is traveling at rate of u_2=1\ m/s

combined velocity of the two is v=0.25\ m/s

Suppose the masses of the clouds be m_1,m_2

Conserving momentum

\Rightarrow m_1u_1+m_2u_2=\left(m_1+m_2\right)v\\\text{Divide whole equation by }m_2\\\Rightarrow \dfrac{m_1}{m_2}u_1+u_2=\left(\dfrac{m_1}{m_2}+1\right)v\\\\\Rightarrow 0+1=0.25\dfrac{m_1}{m_2}+0.25\\\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{0.75}{0.25}\\\\\Rightarrow \dfrac{m_1}{m_2}=3

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$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$

      $=\frac{1}{1.4 \times 10^{-3}} \times 110$

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8 0
3 years ago
jill's car has a maximum acceleration of 8.7 miles per hour per second. how many seconds does it take her to acceleration from 0
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T = ?
v1 = 0mph
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a =  \frac{v2 - v1}{t}
 \\ t =  \frac{v2-v1}{a}
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Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori
LenaWriter [7]

Answer:

Explanation:

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Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

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Vf=15.73m/s

3 0
3 years ago
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I barely do my own homework lol
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