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dsp73
3 years ago
10

To induce an electric current in a coil of a wire you can

Physics
1 answer:
Nata [24]3 years ago
8 0

Answer:

The Answer is

A) move a magnet into the centre of the coil of wire

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The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

  • 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
  • 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0
2 years ago
State 7 branches of physics
erma4kov [3.2K]

Answer:

mechanics

optics

sound

astrophysics

nuclear physics

heat and temperature

electromagnetism

are branches of physics

7 0
3 years ago
Read 2 more answers
Which device is using a motor? A. A water heater creates heat energy from electric energy. B. A waterfall rotates a waterwheel t
Mumz [18]

Answer:

C is the right answer.

Body massager uses electrical energy to move back and forth. In this sense, a motor is being used for the operation

4 0
4 years ago
6. DRAW 3 3/8° LINE<br> 7. DRAW A 5* LINE
Fiesta28 [93]

Answer:

Each part so obtained will represent the fraction 1/8 and the number line obtained will be of the form: To mark 3/8; move three parts on the right-side of zero. To mark 5/8; move five parts on the right-side of zero. To mark -1 3/8 i.e. -11/8; move eleven parts on the left-side of zero.

Explanation:

5 0
4 years ago
How many revolutions per minute would a 25 m -diameter ferris wheel need to make for the passengers to feel "weightless" at the
kozerog [31]
Refer to the diagram shown below.

Let ω =  the angular velocity (rad/s) of the wheel.

At the topmost point, the passenger, with mass m,  will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.

Therefore
mω²r = mg
ω = √(g/r)

Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s

Because 1 revolution per second is 2π rad/s, therefore
\omega = (0.8854 \,  \frac{rad}{s} )*( \frac{1}{2 \pi} \,  \frac{rev}{rad} )*(60 \,  \frac{s}{min} ) = 8.455 \, rev/min

Answer: 8.455 rev/min

8 0
3 years ago
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