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natta225 [31]
2 years ago
5

A spring stretches 0.220 m when a 0.400 kg-mass is hung from it. What is its spring constant? (Mass is not a force )

Physics
2 answers:
Fantom [35]2 years ago
8 0
We want to know the amount of force that stretches the spring 0.22 m.
That force is the WEIGHT of the mass hung from it.
The weight of the mass is (mass) times (gravity).
To do that calculation, we need to know the value of gravity, but
gravity has different values on every planet.  I shall assume that
this whole springy question is taking place on Earth, so that the
value of gravity is 9.8 m/s² .

The weight of the mass is (0.4 kg) x (9.8 m/s²) = 3.92 Newtons.

The spring constant is

(force/length of the stretch)

= (3.92 Newtons) / (0.22 meters)

= (3.92 / 0.22) Newtons/meter

= 17.82 N/m .

romanna [79]2 years ago
8 0

Correct Answer:

17.82 N/m

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Answer:

Explanation:

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It's melting point is 1336 K

It's Heat of fusion is 63000 J/kg

Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,

The first is E1 = 63000 J/kg x 1.5 = 94500 J

the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

Therefore, all solid is not correct. You will have a mixture of solid and liquid.

For more detail, the difference between E1 and E2 is 7812 J, and that will melt

7812/63000 = 0.124 kg of the solid gold

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Answer is given below

Explanation:

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The total charge that an automobile battery can supply without being recharged is given in terms of ampere-hours. A typical 12 V
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7.894 Hours.

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Based on information number hours that this battery will last with give load  has mathematical relation of.

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with load 60A t =  1h, 30A t = 2h so on and forth.

two head lights draw total current of 2x3.8A = 7.6A.

putting this in above relation gives.

t = \frac{60Ah}{7.6A}=7.894 h.

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An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.
viktelen [127]

Answer:

The  force that the wire exerts on the electron is -4.128\times10^{-20}i-6.88\times10^{-20}j+0k

Explanation:

Given that,

Current = 8.60 A

Velocity of electron v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

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B=\dfrac{\mu I}{2\pi d}(-k)

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B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}

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We need to calculate the force that the wire exerts on the electron

Using formula of force

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F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )

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F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k

Hence, The  force that the wire exerts on the electron is -4.128\times10^{-20}i-6.88\times10^{-20}j+0k

5 0
2 years ago
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