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White raven [17]
2 years ago
5

You want to lift a heavy box with a mass L = 56.0 kg using the two-ideal pulley system as shown. With what minimum force do you

have to pull down on the rope in order to lift the box at a constant velocity? One pulley is attached to the ceiling and one to the box.
Physics
1 answer:
Olin [163]2 years ago
3 0

The minimum force required to lift the box at constant velocity is determined as 274.4 N.

<h3>Minimum force required</h3>

The minimum force required to lift the box at constant velocity is the tension in one of the pulleys, and the magnitude is calculated as follows;

2T = mg

where;

  • m is mass of the box
  • T is the minimum force required

2T = mg

T = mg/2

T = (56 x 9.8)/2

T = 274.4 N

Learn more about minimum force here: https://brainly.in/question/47873510

#SPJ1

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When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N

Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
3 years ago
A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.
wolverine [178]

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = m=2.70\times 10^3kg

We know that kinetic energy is given  by

KE=\frac{1}{2}mv^2

So v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec

(B) We know that work done is given by

W = Fd

So F=\frac{W}{d}=\frac{4780}{25.5}=187.45N

4 0
3 years ago
Perform the following calculations and give your answer with the correct number of significant figures
love history [14]

Answer:

see below

Explanation:

a. 0.1886 x 12 =2.2632

This has 2 sig figures so the answer can only have 2 sig figures

2.3

b. 2.995 - 0.16685 =2.82815

The most accurate in the problem is to thousands place so our answer can only be accurate to the thousands place

2.828

c. 910 x 0.18945=172.3995

The  least number of significant figures is 3 so the answer can only have 3 significant figures

172

3 0
3 years ago
The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point
aleksley [76]

The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

6 0
3 years ago
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Answer:

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4 0
2 years ago
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