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Dennis_Churaev [7]
3 years ago
14

20 POINT QUESTION

Physics
2 answers:
Valentin [98]3 years ago
8 0
Label A: sublimation
Label B: condensation
Label C: melting
PIT_PIT [208]3 years ago
4 0

Label A: sublimation

Label B: condensation

Label C: melting

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An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,
maw [93]

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is

43.8 N = <em>k</em> (0.155 m)   ==>   <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

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3 years ago
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Some earthquake waves, a cat’s purr, and elephant communications consist of ultrasonic waves.
Kobotan [32]
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. One long wire carries a current of 30 A along the entire x axis. A second-long wire carries a current of 40 A perpendicular to
d1i1m1o1n [39]

Answer:

magnitude of net magnetic field at given point is

B = 5 \times 10^{-6} T

Explanation:

As we know that magnetic field due to a long current carrying wire is given as

B = \frac{\mu_o i}{2\pi r}

here we we will find the magnetic field due to wire which is along x axis is given as

i = 30 A

r = 2 m

now we have

B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}

B_1 = 3\times 10^{-6} T into the plane

Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as

i = 40 A

r = 2 m

now we have

B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}

B_2 = 4\times 10^{-6} T along + x direction

Since the two magnetic field is perpendicular to each other

So here net magnetic field is given as

B = \sqrt{B_1^2 + B_2^2}

B = 5 \times 10^{-6} T

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