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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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Block m1 of mass 2m and velocity v0 is traveling to the right (+x) and makes an elastic head-on collision with block m2 of mass
Oksana_A [137]
1) In any collision the momentum is conserved

(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')

candel all the m factors (because they appear in all the terms on both sides of the equation)

2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')

2) Elastic collision => conservation of energy

=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2

cancel all the 1/2 and m factors =>

2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>

4(vo)^2 = 2(v1')^2 + (v2')^2

now replace (v2') = -2(v1')

=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>

(v1')^2 = [4/6] (vo)^2 =>

(v1')^2 = [2/3] (vo)^2 =>

(v1') = [√(2/3)]*(vo)

Answer: (v1') = [√(2/3)]*(vo)




 
4 0
3 years ago
What is the magnitude and direction (right or left) of the
Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

Fy_{net}=F_{n}+F_{g} (1)

Where F_{n}=12 N is the Normal force, directed upwards and F_{g}=-12 N is the weight of the box (gravity force), directed downwards.

Fy_{net}=12 N-12 N (2)

Fy_{net}=0 N (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

Fx_{net}=F_{left}+F_{right} (4)

Where F_{left}=-3 N and F_{right}= 15 N

Fx_{net}=-3 N + 15 N (5)

Fx_{net}=12 N (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

5 0
3 years ago
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0
3 years ago
Transcranial magnetic stimulation (TMS) is a noninvasive method for studying brain function, and possibly for treatment as well.
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Answer:

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Explanation:

A = Area = 1.75\times 10^{-2}\ m^2

\dfrac{dB}{dt} = Rate of change of magnetic field = 3\times 10^4\ T/s (assumed)

Induced electromotive force is given by

E=A\dfrac{dB}{dt}\\\Rightarrow E=1.75\times 10^{-2}\times 3\times 10^{4}\\\Rightarrow E=525\ V

The induced electromotive force is 525 V

3 0
3 years ago
Which example describes constant acceleration due only to a change in direction
Degger [83]
A car driving at a constant speed around a circular track
8 0
4 years ago
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