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Morgarella [4.7K]
3 years ago
14

With a pulley system, a force of only 50 lb can lift a 500-lb weight. What is the mechanical advantage if this system is 100% ef

ficient
Physics
1 answer:
saw5 [17]3 years ago
3 0

Answer:

the mechanical advantage is 10

Explanation:

The computation of the mechanical advantage is as follows:

Mechanical advantage is

= Load ÷ Effort

where,

The Loan is equivalent to the weight i.e. 500 lb

And, the effort is equivalent to the force applied  i.e. 50lb

So, the mechanical advantage is

= 500 ÷ 50

= 10

Hence, the mechanical advantage is 10

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What measures we can't answer without the measures
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A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl
frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ =  34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence,  the work done by tension before the block goes up the incline = 292.97 J

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What force would be required to accelerate a 1,100kg car to 0.5 m/s2
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the force required to accelerate a 1,100kg car is 550N

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Why do most amphibians return to the water to reproduce?
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2 years ago
Read 2 more answers
A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal
sukhopar [10]

Answer:

-6112.26  J

Explanation:

The initial kinetic energy, KE_i is given by

KE_i=0.5mv_1^{2} where m is the mass of a body and v_i is the initial velocity

The final kinetic energy, KE_f is given by

KE_f=0.5mv_f^{2} where v_f is the final velocity

Change in kinetic energy, \triangle KE is given by

\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for v_f and 12.6 m/s for v_i and 77 Kg for m we obtain

\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26  J</u>

3 0
3 years ago
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