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ioda
3 years ago
5

Why are electromagnets, rather than permanent magnets, used in MRLs?

Physics
1 answer:
yKpoI14uk [10]3 years ago
5 0

Answer:

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.

Explanation:

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6 0
3 years ago
The temperature of a smelting furnace is found to be 2000 degree Celsius.find the temperature on Fahrenheit scale​
Dima020 [189]

Answer:

3632.

Explanation:

Based on the formula:

2000×(9/5)+32=3632

6 0
2 years ago
When an airplane flies from Dallas - Ft. Worth International airport to Los Angeles International airport, the time in the air i
Elanso [62]

Answer:

7.99 or 8 depends where you round.

Explanation:

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7 0
3 years ago
To which group/family does each of these belong? A. Sulfur _________ B. Sodium _________ C. Argon _________ D. Silicon _________
ArbitrLikvidat [17]

Answer:

A. Sulfur _________ group 16 chalcogens

B. Sodium _________ group 1 alkali metals

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7 0
2 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
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