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Nitella [24]
3 years ago
11

The electric current running through the wire coil in an electric motor exerts force directly onto

Physics
2 answers:
Sveta_85 [38]3 years ago
7 0
C) A powerful magnet
Mashcka [7]3 years ago
6 0

The correct choice is

C) a powerful magnet.

we know that a current carrying wire creates magnetic field around it. inside this magnetic field, when a magnet comes, the magnet experience magnetic force. This same thing happen in motor which is known as motor effect.

the electric current running through coil creates magnetic field which exerts the force on the permanent magnet.


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To test the quality of a tennis ball, you drop it onto a floor from a height of 4.00m . It rebounds to a height of 2.00 m. If th
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3 years ago
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To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 cm makes 621 vibrations per
PilotLPTM [1.2K]

Answer:

T=9.4 N

Explanation:

We are given that

Mass of wire,m=16.5 g=\frac{16.5}{1000}kg

1 kg=1000g

Length of wire,l=75 cm=75\times 10^{-2}m

1 m=100 cm

Wavelength of transverse wave=\lambda=3.33 cm=3.33\times 10^{-2}m

Frequency=621 Hz

Mass per unit length=m_l=\frac{m}{l}=\frac{16.5}{1000\times 75\times 10^{-2}}=0.022 kg/m

\nu=\frac{v}{\lambda}

v=\nu \lambda

Where\nu=frequency of wave

\lambda=Wavelength of wave

Speed of wave=v

Using the formula

v=3.33\times 10^{-2}\times 621=20.7m/s

v=\sqrt{\frac{T}{m_l}}

v^2=\frac{T}{m_l}

T=v^2m_l

Using the formula

T=(20.7)^2\times 0.022=9.4N

Hence, the tension,T=9.4 N

7 0
3 years ago
What causes the moon's phases?
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We can determine the __________ of a wave when given the frequency and the wavelength.
nikklg [1K]
I believe the answer is velocity  
8 0
3 years ago
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A proton is accelerated to 225V. Its de-Broglie wavelength is:
marin [14]

Answer:

The  value  is  \lambda =  1.9109 *10^{-12} \  m

Explanation:

From the question we are told that

  The  potential of the proton is  V  =  225 \  V

Generally the momentum of the particle is mathematically represented as

         p  =  \sqrt{ 2 *  m  *  V  *  e }

Here  e is the charge on the proton with value  

       e =  1.60 *10^{-19} \  C

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So

    p  =  \sqrt{ 2 * (1.67*10^{-27} ) *  225 *  1.6*10^{-19}}

=>    p  = 3.4676 *10^{-22} \  kg \cdot m/ s

So the de-Broglie wavelength isis mathematically represented as

     \lambda  =  \frac{h}{p}

Here  h is the Planck's  constant with value  

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=>   \lambda  =  \frac{6.626 *10^{-34}}{3.4676 *10^{-22} }

=>\lambda =  1.9109 *10^{-12} \  m

   

6 0
3 years ago
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