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2 years ago

In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.

Physics

1 answer:

galben [10]2 years ago

6 0

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED

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1 year ago

A weight lifter applies an upward force of 1100 N while lowering a dumbbell

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Answer:

Explanation:

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3 years ago

Read 2 more answers

A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of

OLEGan [10]

Answer:

c) $F_{net} = 0.640 kN$

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

$F_{net} = ma$

here we know

m = 80 kg

for circular motion acceleration is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{40^2}{200} = 8 m/s^2$

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3 years ago

The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were

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Answer:

a) 45571 N

b) 22786 N

c) 4557 N

Explanation:

Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

$\Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)$

Now, just applying the original form of Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

$\Delta p = F* \Delta t (2)$

So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

$F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)$

$F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)$

$F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)$

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3 years ago

In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.​

In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.