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amm1812
1 year ago
11

A spring with a spring constant value of 2500 startfraction n over m endfraction is compressed 32 cm. A 1. 5-kg rock is placed o

n top of it, then the spring is released. Approximately how high will the rock rise?.
Physics
1 answer:
Sidana [21]1 year ago
7 0

After the spring is released, the rock will rise to 9 m. The result is obtained by using the law of conservation of energy.

<h3>What is law of conservation of energy?</h3>

The law of conservation of energy states that "<em>The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant</em>."

E initial = E final

We have a spring with the spring constant of 2500 N/m. It is compressed 32 cm with a 1.5 kg rock on top of it. Then, the spring is released.

How high will the rock rise?

In that case, there are elastic potential energy and gravitational potential energy.

  • The initial energy is when the spring is compressed. It is elastic potential energy, U = 1/2 kΔx².
  • After the spring is released, the kinetic energy at the maximum height of the rock can rise is zero. The energy left in that situation is gravitational potential energy, U = mgh.

h is the maximum height of the rock can rise.

E initial = E final

1/2 kΔx² = mgh

1/2 × 2500 × (0.32)² = 1.5 × 9.8 × h

1/2 × 2500 × (0.32)² = 1.5 × 9.8 × h

128 = 14.7h

h = 8.71 m

Hence, after the spring is released, the rock will rise to 9 m.

Learn more about law of conservation of energy here:

brainly.com/question/13682185

#SPJ4

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A rectangular steel plate expands as it is heated. Find the rate of change of area with respect to temperature T when the width
tensa zangetsu [6.8K]

Answer:

The variation rate is 5.42 10⁻⁵ cm²/ºC

Explanation:

When we have a thermal expansion problem we must have the relationship of the change in length as a function of the temperature, which are given in this problem, so we can write the expression for the area of ​​a rectangle

      a = L W

They ask us to find the rate of variation of this area depending on the temperature, so we can derive this expression with respect to the temperature

    da / dT = d(LW) / dt

We use the derivative of a product since the two magnitudes change

    da / dT = W dL/dT + L dW/dT

The values ​​they give us are

\frac{dL}{dT} = 1.9 10⁻⁵ cm/ºC

\frac{dW}{dT} = 8.5 10⁻⁶ cm/ºC

W = 1.6  cm

L= 2.8 cm

 Substituting the values ​​and calculating

\frac{da}{dT} = 1.6 1.9 10⁻⁵ + 2.8 8.5 10⁻⁶

\frac{da}{dT} = 3.04 10⁻⁵ + 2.38 10⁻⁵

\frac{da}{dTy}=  5.42 10⁻⁵ cm²/ºC

The variation rate is 5.42 10⁻⁵ cm²/ºC

8 0
4 years ago
A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of
Aneli [31]

Explanation:

Let's solve the problem by using conservation of momentum:

m1 u1 + m2 u2 = m1 v1 + m2 v2

where:

m1 = 3.5 g  is the mass of the blue marble

m2= 1.2 g is the mass of the red marble

u1 = 15 cm/s  is the initial velocity of the blue marble

u2 = 3.5 cm/s is the initial velocity of the red marble

v1 = 5.5 cm/s is the final velocity of the blue marble

We can find the final velocity of the red marble by re-arranging the equation and solving for v2 :

v2 = 1/m²(m1 u1 + m2 u2 - m1 v1)=

=1/1.2(3.5×15+1.2×3.5−3.5×5.5 )=31cm/s

7 0
4 years ago
What does a measured number tell you?
oksano4ka [1.4K]

Answer:

The precision at which the number was measured  

Explanation:

The number of digits that you use to record a measurement indicates the precision of the measurement.

5 0
4 years ago
A flat screen is located 0.55 m away from a single slit. Light with a wavelength of 544 nm (in vacuum) shines through the slit a
kobusy [5.1K]

Answer:

Explanation:

The diffraction pattern is given as

Sinθ = mλ/ω

Where m=1,2,3,4....

Now, when m=1

Sinθ = λ/ω

Then,

ω = λ/Sinθ

The width of the central bright fringe is given as

y=2Ltanθ. From trigonometric

Then,

θ=arctan(y/2L)

Given that,

y=0.052m

L=0.55m

θ=arctan(0.052/2×0.55)

θ=arctan(0.0473)

θ=2.71°

Substituting this into

ω = λ/Sinθ

Since λ=544nm=544×10^-9m

Then,

ω = 544×10^-9/Si.2.71

ω = 1.15×10^-5m.

5 0
3 years ago
60 POINT!!!!!!
Genrish500 [490]
You in college stop cheating
8 0
3 years ago
Read 2 more answers
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