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Dennis_Churaev [7]
3 years ago
8

What happens when the clapper of an electric doorbell move to strike the bell?

Physics
1 answer:
faust18 [17]3 years ago
5 0

Answer: The clapper of an electric doorbell is an electromagnet. When the clapper of a doorbell strikes the bell, it opens an electric circuit. The electromagnet of an electric motor is connected to a permanent magnet. Only the shaft of an electric motor turns when current flows through the motor.

Explanation: hope this helped

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A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What
kati45 [8]

Explanation:

The given data is as follows.

           radius (r) = 3.25 cm,    \alpha = 11.6 rad/s^{2}

Now, we will calculate the tangential acceleration as follows.

          a_{tangential} = \alpha \times r

Putting the given values into the above formula as follows.

         a_{tangential} = \alpha \times r

                      = 11.6 rad/s^{2} \times 3.25 cm

                      = 37.7 rad cm/s^{2}

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 rad cm/s^{2}.

6 0
3 years ago
Just like energy is <br> matter is lost through an ecosystem
Novosadov [1.4K]
Yes energy is reduced 
7 0
2 years ago
Read 2 more answers
Look at figure D . Explain what will happen if a larger mass it put on the force meter
aev [14]

it will experience great force

5 0
3 years ago
A straight segment of wire 35.0 cm long carrying a current of 2.60 A is in a uniform magnetic field. The segment makes an angle
den301095 [7]

Answer:

Magnetic field, B = 0.275 T

Explanation:

Given that,

Length of the wire, L = 35 cm = 0.35 m

Current carried in the wire, I = 2.6 A

The segment makes an angle of 53∘ with the direction of the magnetic field, \theta=53^{\circ}

Magnetic force, F = 0.2 N

To find,

The magnitude of the magnetic field.

Solution,

The magnetic force acting on the wire is given by :

F=ILBsin\theta

\theta is the angle between the length of wire and the magnetic field.

0.2=2.6\times 0.35\times B\times sin(53)

B = 0.275 T

Therefore, the magnitude of the magnetic field is 0.275 T. Hence, this is the required solution.

5 0
2 years ago
A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase
SashulF [63]

Answer:

E_Phase = 560V

Explanation:

The computation of the voltage i.e. dropped across each phase is shown below:

Given that

The delta connection line voltage is

E_line = 560 V

And, in the case of delta connection, the line voltage would be equivalent to the phase voltage

That means

E_Phase = E_Line

= 560 V

Hence, the voltage i.e. dropped across each phase is

E_Phase = 560V

5 0
3 years ago
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