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HACTEHA [7]
3 years ago
9

As a rocket rises, its kinetic energy changes

Physics
2 answers:
djverab [1.8K]3 years ago
5 0
The answer is True
hope i helped
Bond [772]3 years ago
3 0
<span>Transformed into potential energy</span>
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Oxygen enters the blood Through your lungs
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Which of the following describes the electron sharing between hydrogen and fluorine A.Hydrogen and fluorine share one electron w
atroni [7]
Hydrogen has one electron in its outermost shell, while fluorine has seven electron in its outermost shell, hence both hydrogen and fluorine needs a single electron to complete its outermost shell. 
That's why there is a single bond between hydrogen and fluorine.
Hence both hydrogen and fluorine share one electron with each other, so option "A" is correct.
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3 years ago
Hi I’m having some difficulty with this question could really use some help!
Serhud [2]
\begin{gathered} v=\lambda f \\ v=4m\cdot2s^{-1} \\ v=8m/s \end{gathered}

3 0
1 year ago
11. A plow pushes 100 kg of snow with 300 N of force. How much is the pile of snow<br> accelerated?
Fed [463]

Answer:

Explanation:

This is an application of Newton's second Law.

Formula

F = m * a

F = 300 N

m = 100 kg

a = ?

F = m * a

300N = 100 kg * a            Divide by 100

300N/100kg = a

a = 3 m/sec^2

3 0
2 years ago
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
3 years ago
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