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artcher [175]
3 years ago
10

If a system requires 150 j of input work and produces 123 J of output work, whats its effiency

Physics
1 answer:
Yuki888 [10]3 years ago
4 0

The effiency of a machine is

                 (output work or energy) / (input work or energy) .

For the system described in the question, that's

                 (123 J) / (150 J) = 0.82  =  82% .

 
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Answer:Dalton just expanded on the Greek idea of the atom.

Explanation:

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Answer:

F=G(m1m2)/Rsquare if radius is given

F=G(m1m2)/dsquare if distance is given

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f =gravitational force

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Two protons (each with q = 1.60 x 10-19)
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Answer:

230.4 N

Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

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The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

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5 0
2 years ago
A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an
Roman55 [17]

Answer:

v_{2f} = \frac{2vm_1}{m_2 + m_1}

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

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P_i = m_1v

After the collision

P_f = m_1v_{1f} + m_2v_{2f}

So using the law of momentum conservation

P_i = P_f

m_1v = m_1v_{1f} + m_2v_{2f}

We can solve for the speed of ball 1 post collision in terms of others:

v_{1f} = v - v_{2f}\frac{m_2}{m_1}

Their kinetic energy is also conserved before and after collision

m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2

m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2

From here we can plug in v_{1f} = v - v_{2f}\frac{m_2}{m_1}

m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2

m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2

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v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0

v_{2f}(1 + \frac{m_2}{m_1}) = 2v

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8 0
3 years ago
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crimeas [40]

Answer:

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Explanation:

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