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mash [69]
2 years ago
8

A spring is hung from the ceiling. A 2.0-kg mass suspended hung from the spring extends it by 6.0 cm. A downward external force

applied to the mass extends the spring an additional 10 cm. What is the work done by the force
Physics
1 answer:
Stolb23 [73]2 years ago
8 0

The work done by force on a spring hung from the ceiling will be 1.67 J

Any two things with mass are drawn together by the gravitational pull. We refer to the gravitational force as attractive because it consistently seeks to draw masses together rather than pushing them apart.

Given that a spring is hung from the ceiling with a 2.0-kg mass suspended hung from the spring extends it by 6.0 cm and a downward external force applied to the mass extends the spring an additional 10 cm.

We need to find the work done by the force

Given mass is of 2 kg

So let,

F = 2 kg

x = 0.1 m

Stiffness of spring = k = F/x

k = 20/0.006 = 333 n/m

Now the formula to find the work done by force will be as follow:

Workdone = W = 0.5kx²

W = 0.5 x 333 x 0.1²

W = 1.67 J

Hence the work done by force on a spring hung from the ceiling will be 1.67 J

Learn more about force here:

brainly.com/question/12970081

#SPJ4

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Car's mass is 950kg. It's driven at 100km/h. How big is the kinetic energy?
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Answer:

4750000 J

Explanation:

Kinetic Energy= 1/2* mass* velocity²

1/2*950*100²=4750000

8 0
3 years ago
1 jWhat is the correct unit for temperature? <br>​
Nadya [2.5K]

Answer:

Answer: Kelvin ________________

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3 years ago
A stone is dropped from rest from the top of a building. It takes Δt = 2.2 s for it to reach the ground.
NeTakaya

Answer:

Value of magnitude of acceleration will be 9.8m/sec^2

Explanation:

It is given that when a stone is dropped it takes 2.2 sec to reach the ground

(a) As the stone is dropped from the top of building

So its initial velocity u_i will be 0 m /sec

(b) As the stone is free falling and there is no external force applied on it so its acceleration will be equal to acceleration due to gravity

So value of magnitude of acceleration will be equal to 9.8m/sec^2

6 0
3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
4 years ago
PLEASE HELPPP!!! I'LL GIVE BRAINLEST FOR CORRECT ANSWERS​
babymother [125]

Answer:

Explanation:

Weight will be highest at the pole where there is no tangential velocity requiring centripetal acceleration. Also, due to the slight bulge of the earth at the equator, the distance from the surface to the center of mass of earth is slightly shorter there meaning gravity is slightly stronger. Fg = GmM/R²

Weight will be lowest at the equator where the object is moving about the earth axis at an angular velocity of 2π/(24(3600)) rad/s ( about 7.27e-5 rad/s)

This means that some of the (already weaker, see above) gravity there is required to supply the needed centripetal acceleration to keep the object on the ground.

4 0
3 years ago
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