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2 years ago

Suppose the number of coils in the solenoid of an electromagnet is increased. What happens to the electromagnet?

Physics

2 answers:

Akimi4 [234]2 years ago

5 0

Answer:

Its strength becomes greater.

Explanation:

vagabundo [1.1K]2 years ago

4 0

Answer:

Answer is Below

Explanation:

According to a Quizlet, An electromagnet is a solenoid with a ferromagnetic core. Both a solenoids and an electromagnet's magnetic field increases when the number of coils, closeness of the coils, and current increases.

Hopes this Helps :D

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Which energy transfer occurs when a cube of ice is placed in a glass of<br> water?

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Acetic acid is a weak acid and sodium hydroxide is strong base. Salts of the two will hydrolyse to give basic solution. So, at neutral point, pH of the solution will be greater than 8.

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2 years ago

If an astronaut can throw a certain wrench 15.0 m vertically upward on earth, how high could he throw it on our moon if he gives

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Answer:

hm= 18.2 m

Explanation:

Solution is attached

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3 years ago

Light is an electromagnetic wave and travels at the speed of 3.00 x 10^8 m/s

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3 years ago

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

$f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\$

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

$f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz$

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

$f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz$

3 0

2 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago