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Vaselesa [24]
2 years ago
14

How much weight can be placed on the spring so that the end of the spring is 2m above the ground?

Physics
1 answer:
Ronch [10]2 years ago
8 0

The weight must be 288N placed on the end of the spring when it is 2m from the ground.

<h3>What is gravitational potential energy?</h3>

If an object is lifted, work is done against gravitational force. The object gains energy.

Given is a spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground. The final height of weight above the ground, h =2m

Using Hooke's law, expressed as

Force, F = k (x₂-x₁).

Put the values, we get

F = 48 (8-2)

F = 288 N

This force applied on the spring is equal to the weight placed on the spring.

Weight W= 288N.

Thus, the weight must be 288N.

Learn more about gravitational potential energy.

brainly.com/question/3884855

#SPJ1

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3 years ago
Neptune is approximately 4.5 billion kilometers from the sun. What is this distance in AU?
asambeis [7]
Well, one AU is 149,597,870 km. So, we would basically have to divide 4.5 billion km by 149,597,870, right?

4,500,000,000/149,597,870=30.080642 AU.

So, the correct answer would be 30 AU. Hoped this helped!
7 0
3 years ago
Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
An acorn with a mass of .0300 kg is hanging from a branch in a tree it is 2.50 M off the ground what is the potential energy of
Alekssandra [29.7K]
According to another source this is what I got
<span>0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m ) 
</span>Hope it helps
5 0
3 years ago
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What is the difference between a free body diagram and a vector diagram?​
Murrr4er [49]

Answer:

Free body diagrams are used to describe situations where several forces act on an object. On the other hand Vector diagrams are used to resolve (break down) a single force into two forces acting as right angles to eachother

Explanation:

Hope this helps !

4 0
2 years ago
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