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Andre45 [30]
3 years ago
15

A small circular loop of 5 mm radius is placed 1 m away from a 60-Hz power line. The voltage induced on this loop is measured at

0.6 µV. What is the current on the power line

Physics
1 answer:
marin [14]3 years ago
5 0

Answer:

i = 101.4A

Explanation:

The steps to the solution can be found in the attachment below.

We have been given the frequency f = 60Hz. From this we can calculate the angular frequency of the power line.

It is assumed that sinwt = –1 in order to calculate the time varying current. Although the magnitude of the current is large, consideration should also be given to the distance between the coil and the power line. The induced emf is small considering the area of the coil which is 7.85×10-⁵m².

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Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, P_{atm} = 101 Kpa

P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}

P_{gauge}  = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

P_{gauge}  = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge}  = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge}  = 110,298.26 pa

Therefore, the gauge pressure of air is 110 kpa

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mixas84 [53]

The force is still 1 pound.

The TORQUE around the pivot is (force) x (distance from the pivot) = 1 foot-pound. (B)

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Rudiy27

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