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Elanso [62]
4 years ago
12

A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

speed of 0.76 m/s when it leaves the chute. Determine the speed of the car–coal system after the coal has come to rest in the car.
Physics
1 answer:
Nady [450]4 years ago
8 0

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

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Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}

Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

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lukranit [14]

Answer:

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Explanation:

Conversion

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5 0
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Newton's Third Law states that every action has an equal and opposite reaction.

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tan θ = vᵧ / vₓ

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