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kakasveta [241]
1 year ago
5

A current carrying wire of length 50cm is positioned perpendicular to a uniform magnetic field

Physics
1 answer:
RideAnS [48]1 year ago
3 0

A current-carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field. The magnetic field strength is 0.6 Tesla.

<h3>What is magnetic field?</h3>

The magnetic field is the region of space where an object experiences a magnetic force as it enters the field.

Given is the wire of length 50 cm = 0.5 m and the current is 10.0 A. There is a resultant force of 3.0 N on the wire due to the interaction of the current and field.

The relation between magnetic field strength and current is

F = ILB

Substituting 50 for L, 10 for I and 3 for F, we get the magnetic field strength B.

3 = 10 × 0.5 × B

B= 0.6 Tesla

Therefore, the magnetic field strength is 0.6 Tesla.

Learn more about magnetic field.

#SPJ

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How can you double the frequency of a wave if you have control over both the wavelength and the wave velocity?
sammy [17]

Answer:

Change the wavelength to half while keeping velocity constant or change the wave velocity to 2 times  while keeping wavelength constant.

Explanation:

The frequency of a wave can be defined as the rate of change of wave speed with respect to the wavelength of the wave.

Mathematically,

f=\frac{v}{\lambda}

Here, v is the velocity, lambda is the wavelength, and f is the frequency of the wave.

If the observer want to double the frequency of the wave, then he should increase the wave velocity two times and take wavelength constant for this case or either he should half the wavelength of the wave to take the velocity of the wave constant.

5 0
3 years ago
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How much power is required to do 180 J of work in 2.4s?​
rjkz [21]
75
p = w / t
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help plz! what vibrates in following types of wave motion 1)light wave 2)sound waves 3)x-rays 4)water waves​
Alecsey [184]

Answer:

I believe it's 2) sound waves

Explanation:

With sound waves, the energy travels along in the same direction as the particles vibrate. This type of wave is known as a longitudinal wave, so named because the energy travels along the direction of vibration of the particles.

7 0
2 years ago
If an atom has 15 protons, 12 neutrons, and 17 electrons, what is the atom's electrical charge? A. -5 B. -2 C. +3 D. +5
Nimfa-mama [501]

Answer:

B. -2

Explanation:

The total charge on an atom is the sum of all individual charges present in it. Therefore, the total charge on this atom is given by the following formula:

q = (n_p)(q_p)+(n_e)(q_e)+(n_n)(q_n)

where,

q = total charge on atom = ?

n_p = no. of protons in the atom = 15

n_e = no. of electrons in the atom = 17

n_n = no. of neutrons in the atom = 12

q_p = charge on proton = +1

q_e = charge on electron = -1

q_n = charge on neutron = 0

Therefore,

q = (15)(1)+(17)(-1)+(12)(0)\\q=-2

Hence the correct option is:

<u>B. -2</u>

5 0
3 years ago
From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ
Alexandra [31]

Approximately 15 m/s is the speed of the car.

<u>Explanation:</u>

<u>Given:</u>

speed of sound - 343 m/s

You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,

f^{\prime}=0.959 f

\frac{f^{\prime}}{f}=0.959

If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,

                     f^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) \times f

Where,

f^' - Observed frequency

f – Actual frequency

v – Velocity of sound waves

v_0 – Velocity of observer

v_s - velocity of source

When source moves away from an observer at rest (v_{0} = 0), the equation would be

                        f^{\prime}=\left(\frac{v}{v-\left(-v_{s}\right)}\right) \times f

                        f^{\prime}=\left(\frac{v}{v+v_{s}}\right) \times f

                        \frac{f^{\prime}}{f}=\left(\frac{v}{v+v_{s}}\right)

By substituting the known values, we get

            0.959=\left(\frac{343}{343+v_{s}}\right)

           0.959\left(343+v_{s}\right)=343

           0.959(343)+0.959\left(v_{s}\right)=343

           328.937+0.959 v_{s}=343

           0.959 v_{s}=343-328.937=14.063

           v_{s}=\frac{14.063}{0.959}=14.66 \mathrm{m} / \mathrm{s}

Approximately 15 m/s is the speed of the car.

3 0
3 years ago
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