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AVprozaik [17]
4 years ago
9

A boy kicks a ball with a force of 40 N. At exactly the same moment, a gust of wind blows in the opposite direction of the kick

with a force of 40 N. What happened to the ball?
A. It began to move with a speed of 80 m/s.

B. It did not move.

C. It moved in the direction that the wind was blowing.

D. It moved in the direction of the kick.
Chemistry
2 answers:
aliya0001 [1]4 years ago
8 0
B, the opposing forces are the same, thus, the ball doesn't move back or forward.
a_sh-v [17]4 years ago
7 0

The answer is B the ball doesn’t move

40n—>  <— 40n it equals out

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The reaction is shown in the image.

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How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin?
Elena-2011 [213]
3 moles of oxygen will react with 1 mole of ethylene. Convert 12.9 L of oxygen to x moles of oxygen, then divide by three.
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What mass of oxygen is needed for the complete combustion of 4.60×10−3g of methane?
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Then using dimension analysis: 

4.60*10^{-3} g CH4 ( \frac{moleCH4}{16 g CH4}) * ( \frac{2mole O2}{mole CH4}) * ( \frac{32 g O2}{mole O2} ) =  0.0184 g O_{2}
7 0
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A change in 1 degree Celsius is = a change in 1 Kelvin<br> True<br> False
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Answer:

true

Explanation:

because I'm smart yes yes

5 0
3 years ago
What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
sammy [17]

Explanation:

It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

8 0
4 years ago
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