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lukranit [14]
2 years ago
11

If a machine has 100% efficiency, it means that

Physics
2 answers:
I am Lyosha [343]2 years ago
5 0
The answer should be A, because efficiency = (effective work)/(executed work) * 100%
Tresset [83]2 years ago
3 0

Answer: (A).Output work is equal to input work.

Explanation:

Efficiency: Efficiency is the ability of machine.

Efficiency is the maximum of the machine that means the machine gives the maximum output with minimum input.

So we can say that,

If a machine has 100% efficiency, it means that the machine gives output equal to the input.

Hence, Output work is equal to input work.

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Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
wlad13 [49]
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

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3 years ago
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