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kompoz [17]
3 years ago
9

A 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.8 m/s left. Afterwards, the 282 kg car m

oves right at 0.800 m/s. What is the momentum of the 155 kg car afterwards? (Unit=kg*m/s)

Physics
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

482.4kgm/s

Explanation:

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Answer:

40N in either direction is the answer

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3 years ago
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gogolik [260]

Answer:

What’s your question

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3 years ago
A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber
Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}

Now calculate the bouyant force acting on the lead

F_L = Vpg

F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N

This force will act in upward direction

Gravitational force on the lead due to its mass  will act in downward direction

Hence the difference of this two force

T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N

If V is the volume submerged in the water then bouyant force on the bobber is

F_B=V'pg

Equate bouyant force with the tension and gravitational force

F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}

Now Total volume of bobble is

\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\

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Answer:

Definitely Spinning permanent magnets within an array of fixed permanent magnets

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Any relative motion between magnets (be they permanent or electromagnetic) and a coil of wire will induce an electric current in the coil.

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<em>Therefore, spinning permanent magnets within an array of fixed permanent magnets does not induce an electric current.</em>

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2 years ago
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Determine the net work a hiker must do on a 3.35-kg backpack to carry it up a
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Answer:

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