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Ede4ka [16]
3 years ago
15

PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton b

ox 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Physics
1 answer:
GuDViN [60]3 years ago
3 0

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

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Answer: 100 units

Explanation:

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3 years ago
A factory worker pushes a 31.0 kg crate a distance of 4.20 m along a level floor by pushing downward at an angle of 32.0 ∘ below
Darina [25.2K]

Answer:

W_{fr} = 281.539\,J

Explanation:

The kinetic force of friction is:

f = (0.260)\cdot (31\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot \cos 32^{\textdegree}

f = 67.033\,N

Before calculating work, it is require to determine if force exerted on the crate is enough to move it at least. The equation of equilibrium for the crate is:

\Sigma F = F - f + m\cdot g \cdot \sin \theta = m \cdot a

The acceleration experimented by the crate is:

a = \frac{F-f}{m} + g\cdot \sin \theta

a = \frac{118\,N-67.033\,N}{31\,kg}+(9.807\,\frac{kg}{m^{2}} )\cdot \sin 32^{\textdegree}

a = 6.841\,\frac{m}{s^{2}}

This positive result indicates that motion is physically reasonable. Hence, the work done by the force of friction is:

W_{fr} = (67.033\,N)\cdot (4.20\,m)

W_{fr} = 281.539\,J

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4 years ago
The electromagnetic spectrum includes radio waves inferared waves and x rays
olya-2409 [2.1K]

Answer:

True

Explanation:

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3 years ago
Newton’s law of universal gravitation stars that every object in the universe attracts every other object
LenaWriter [7]

Yes it does !  Uh huh. Right you are. Truer words are seldom written.

You have quoted the law quite accurately but also incompletely.  

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4 0
3 years ago
Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an
Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed  s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance  d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1    .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 )     ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0
3 years ago
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