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Law Incorporation [45]

2 years ago

14

What is the magnetic force of a 30m long power line that carries a current of 50A. The magnetic field runs perpendicular to the

power line and has a strength of 30T.
A. 45000 T/N

B. 45 V

C. 45 N

D. 45000 N

1 answer:

Ne4ueva [31]2 years ago

8 0

Answer:

F=BILsin90 when perpendicular sin90 =1 30T x50x30 so you can get 45000N

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What is the student's kinetic energy at the bottom of the hill if he is moving

soldi70 [24.7K]

Answer:

KE = 10530 J or 10.53 KJ

Explanation:

The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

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2 years ago

A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m

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Answer:

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3 years ago

Suppose the force of gravity on the electron was comparable to the electric force created by the deflection plates. how would th

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"F=Vector Sum Of The Two Forces" Is the answer.

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2 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

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2 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a

timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack ( $v$ ), measured in meters per second, is equal to the product of the angular speed of the wheel ( $\omega$ ), measured in radians per second, and the distance of the tack respect to the rotation axis ( $R$ ), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:

$\omega = 2.83\,\frac{rev}{s} \times \frac{2\pi\,rad}{1\,rev}$

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Then, the tangential speed of the tack is: ( $\omega \approx 17.781\,\frac{rad}{s}$ , $R = 0.393\,m$ )

$v = \left(17.781\,\frac{rad}{s} \right)\cdot (0.393\,m)$

$v = 6.988\,\frac{m}{s}$

The tangential speed of the tack is 6.988 meters per second.

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2 years ago