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2 years ago

15

A power cable carries a 100 A current. At what distance from the wire is the magnetic field equal to that of earth, 5.00 x 10^-5

T?

1 answer:

jek_recluse [69]2 years ago

3 0

Answer:

0.4 m

Explanation:

The magnetic field strength is given by the formula ...

B = (μ₀I)/(2πr)

where I is the current in the wire in amperes, and r is the distance from the wire in meters. The value of μ₀ is 4π×10^-7. We want to find r for B = 5×10^-5. This gives ...

5×10^-5 = (4π×10^-7)(100)/(2πr)

r = 2×10^-5/(5×10^-5) = 0.4

At a distance of 0.4 meters from the wire, the magnetic field is equal to that of Earth.

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3 years ago

Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the botto

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Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle ( $v_{bottom}\approx v_{top}$ ):

$p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h$ (1)

$p_{bottom}$ , $p_{top}$ - Total pressures at the bottom and at the top, measured in pascals.

$\rho$ - Density of the water, measured in kilograms per cubic meter.

$\Delta h$ - Height difference of the step, measured in meters.

If we know that $p_{bottom} = 132000\,Pa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 0.275\,m$ , then the pressure at the top of the step is:

$p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h$

$p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)$

$p_{top} = 129303.075\,Pa$

$p_{top} = 129.303\,kPa$

The pressure at the top of the step is 129.303 kilopascals.

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3 years ago

What are 3 things that can affect gravity?

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The force of gravity the masses exert on each other. If one of the masses is doubled , the force of gravity between the objects is doubled. Increases , the force of gravity decreases.

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3 years ago

What does a cell use to eliminate a substance that is too large to leave by diffusion?

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<span>a cell eliminates endocytosis.

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4 years ago

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

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Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

$m_sv_s + m_rv_r = 0$

where $m_s = 72.8, m_r = 265$ are the mass of the swimmer and raft, respectively. $v_s = 5.21 m/s, v_r$ are the velocities of the swimmer and the raft after the run, respectively. We can solve for $v_r$

$265v_r + 72.8*5.21 = 0$

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So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

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3 years ago