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Leto [7]
3 years ago
5

If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil

l
A decrease.
B increase.
C increase, then decrease.
D not change.
Physics
2 answers:
mina [271]3 years ago
6 0
The answer is A.) decrease.
Take pushing a box for example. You push your hardest but then give out, still trying to push. You're doing less work than what you started with.
Dominik [7]3 years ago
6 0

Answer:

Option A is the correct answer.

Explanation:

We know that the expression for work

          Work = Force x Displacement

So work is directly proportional to force.

Here force is decreasing, so work also decreases.

Option A is the correct answer.

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A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0
2 years ago
(D.) Define the following Moment (.) Moment of a force (.) couple of force (.) torque<br>​
erik [133]
Moment of force. Hope this helped
4 0
3 years ago
Two objects with a mass of 10,000-kg each and a distance of 1 meter between them.
Fofino [41]

Answer:

d) What is the force if we doubled both the masses AND we doubled the distance

4 0
3 years ago
The universal law of gravity states that all objects attracts all other objects.
andreev551 [17]
When the object is big enough to contract itself into a ball.
4 0
3 years ago
A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
3 years ago
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