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An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h

Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

$F=\mu mg$

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

$F = 0.77\times85\times9.8$

$F= 641.41\ N$

We need to calculate the acceleration

Using formula of newton's second law

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{ 641.41}{109000}$

$a=0.0059\ m/s^2$

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

3 0

3 years ago

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

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7 0

2 years ago

A motor does 8000j of work in 20 seconds. What is the power of the motor

lozanna [386]

Power = work / time = 8000J / 20s = 400W

5 0

3 years ago

What is angle of dip

DENIUS [597]

<em>Angle of Dip is the angle in the vertical plane aligned with magnetic north (the magnetic meridian) between the local magnetic field and the horizontal.</em>

8 0

3 years ago

A mother and her young child want to play on a seesaw at a playground. The child sits on the end of one side of the seesaw. Wher

EleoNora [17]

Answer:middle

Explanation:

Because it will make the seasaw balanced

4 0

3 years ago