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rosijanka [135]
3 years ago
15

A pulley is used to lift a heavy mass from the floor to the platform. The IMA of the system is 5 and the AMA is 2.5. What is the

efficiency of the system?
Physics
2 answers:
Nuetrik [128]3 years ago
4 0

Answer:

efficiency=\frac{AMA}{IMA}=\frac{2.5}{5}=0.5 or 50%

Explanation:

IMA means Ideal Mechanical Advantage, and it indicates the ratio between the input forces and the output forces without taking into consideration the force needed to overcome friction. In pulleys, this is calculated by counting the number of ropes on the system that in this case is 5

IMA = 5

IMA is greater to the AMA because it doesn't take into consideration friction.

AMA is the Actual Mechanical Advantage, and it is the measure of the usefulness of the pulley, and it indicate the ratio between the output force (Fo) and the input force (Fi). The smaller this ratio, the less useful is the pulley, meaning, it does not simplify the job it is designed to ease, for example, a pulley that does not reduce the force needed to lift a load.

AMA=2.5

So from these two definitions, if we have the actual force ratio and the ideal force ratio, we basically have the "what it is" and the "what could be", a very crude representation of efficiency:

efficiency=\frac{AMA}{IMA}=\frac{2.5}{5}=0.5 or 50%

NemiM [27]3 years ago
3 0
50% as it’s just Ama/ima simple
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3 years ago
James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a
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Answer:

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Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\

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Now we can find the velocity after 2 seconds.

v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]

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Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\

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Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

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8 0
3 years ago
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