2 years ago

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Physics

1 answer:

ss7ja [257]2 years ago

7 0

Answer:

Bro where is image Hope you understand me

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3 years ago

Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a

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Answer:

Acceleration of the second particle at that moment is given as

$a_2 = 2.12 m/s^2$

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

$F = 0 = m_1a_1 + m_2a_2$

now we have

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$m_2 = 0.250 kg$

$a_1 = 2.65 m/s^2$

now we have

$0.200(2.65) + 0.250a_2 = 0$

so we have

$a_2 = 2.12 m/s^2$

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3 years ago

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Explanation:

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∑F = ma

∑F = (5.0 kg) (2.0 m/s²)

∑F = 10 N

b. The net force is the sum of the individual forces.

10 N = F − 5 N

F = 15 N

c. Friction force here is mgμ.

mgμ = 5 N

(5.0 kg) (10 m/s) μ = 5 N

μ = 0.1

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