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likoan [24]
3 years ago
7

What is one advantage of using electromagnets instead of permanent magnets

Physics
2 answers:
zimovet [89]3 years ago
5 0
The main advantage of an electromagnet over a permanent magnet is that the magnetic field can be quickly changed by controlling the amount of electric current in the winding. However, unlike a permanent magnet that needs no power, an electromagnet requires a continuous supply of current to maintain the magnetic field.
olganol [36]3 years ago
4 0
The main advantage of an electromagnet over a permanent magnet is that the magnetic field can be quickly changed by controlling the amount of electric current in the winding. However, unlike a permanent magnet that needs no power, an electromagnet requires a continuous supply of current to maintain the magnetic field.
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Valentin [98]
Diagram A is the correct diagram
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3 years ago
The fastest possible conduction velocity of action potentials is observed in ________.
irga5000 [103]
Thick, myelinated neurons
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3 years ago
Correlation is:
xz_007 [3.2K]

Answer: The answer is B

Explanation: A correlation generally is a mutual relationship/connection or the process of establishing the relationship/connection between 2+ things

6 0
2 years ago
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"
Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0
3 years ago
A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for
emmasim [6.3K]

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

4 0
3 years ago
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