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ElenaW [278]
3 years ago
12

What are some ways that electric fields are similar to magnetic fields?

Physics
1 answer:
sattari [20]3 years ago
3 0

Answer:

The correct answers are the proportionality of the fields concerning distance, vector fields, and forces at a distance.

Explanation:

The similarities between magnetic fields and electric fields are that electric fields are produced by two charges that can be positive and negative. Magnetic fields are associated with two magnetic poles, although they are also produced by moving charges. Both fields are inversely proportional to the square of the distance between the sources, both fields are vectorial and both act by distant forces.

Have a nice day!

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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
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Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

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y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

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t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

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V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

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Step 2:  Finding V_{o}:

From equation(4)

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5 0
3 years ago
Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
ella [17]

Answer:

\frac{dR(t)}{dt}=0.06\Omega

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Deriving what is left (remember that (\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)):

\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}

So we have:

\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

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