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____ [38]
3 years ago
9

An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular

value at this point. Later, she must move the sample to a 7.0 cm distance, but she would like to keep the field the same. By what factor must she increase the current?
Physics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

she must increase the current by factor of 7

Explanation:

The magnetic field produced by a steady current flowing in a very long straight wire encircles the wire.In order to solve the question, we use this formula,

B= μo I/(2πr)

where,

'μo'  represents permeability of free space i.e 4π*10-7 N/A2

B=magnetic field

I= current

r=radius

->When r= 1cm=> 0.01m

B1 = μo I_{1/(2π x 0.01)

->when r=7cm =>0.07m

B2 = μo I_{2}/(2π x 0.07)

Now equating both of the magnetic fields, we have

B1= B2

μo I_{1/(2π x 0.01)= μo I_{2}/(2π x 0.07)

I_{1/  I_{2}= 0.01/0.07

I_{1/  I_{2}= 1/ 7

Therefore, she must increase the current by factor of 7

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Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

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                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

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3 0
3 years ago
A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit
Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

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Where,

a = acceleration

t = time

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y_0 = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

v = \frac{x}{t}

x = Displacement

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With the data we have we can find the time as well

v = \frac{x}{t}

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t = \frac{18.6}{34}

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y = \frac{1}{2} at^2+v_0t+y_0

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3 years ago
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