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sveta [45]
3 years ago
11

A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis

tance 0.0600 m below the original position, released, and allowed to oscillate. How high above the floor is the mass at the highest point in its oscillation
Physics
1 answer:
den301095 [7]3 years ago
3 0

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

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Answer:

C) W = - 190 J

Explanation:

Notation

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6 0
3 years ago
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Answer:

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Explanation:

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6 0
2 years ago
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal t
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Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

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This can be written mathematically as;

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(\frac{\delta \phi _B}{\delta t} ) is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

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ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = \frac{\delta \phi _B}{\delta t} = EMF

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8 0
3 years ago
Explain why a moving object cannot come to a stop instantaneously (in zero seconds)
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Explanation:

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7 0
3 years ago
A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 tim
stellarik [79]

Answer:

T2=336K

Explanation:

Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:

where:

In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)

p1 and p2 are the vapour pressures at temperatures 

T1 and T2

ΔvapH = the enthalpy  of vaporization of the liquid

R = the Universal Gas Constant

p1=p1, T1=307K

p2=3.50p1; T2=?

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R=8.314J.K^-1moL^-1

In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)

P1 and P1 cancelled out:

In(3.50)=4511.667(T2 - 307/307T2)

1.253=14.696(T2 - 307/T2)

1.253=(14.696T2) - (14.696*307)/T2

1.253T2=14.696T2 - 4511.672

Therefore,

4511.672=14.696T2 - 1.253T2

4511.672=13.443T2

So therefore, T2=4511.672/13.443=335.61

Approximately, T2=336K

6 0
3 years ago
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