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nydimaria [60]
3 years ago
14

A long solenoid that has 1,140 turns uniformly distributed over a length of 0.350 m produces a magnetic field of magnitude 1.00

X 10^-4 T at its center. What current is required in the windings for that to occur? Apply the expression for the magnetic field along the axis of a solenoid to find the current.
Physics
1 answer:
Troyanec [42]3 years ago
3 0

Answer:

I = 2.4 *10^{-2} A = 2.4 mA

Explanation:

The magnetic field B inside long solenoid with current I is given as

B = \frac{N \mu_o I}{L}

where

N is number of turn of solenoids = 1140 turns

\mu_0 = 4*\pi *10^{-7} T.m

I is current that passes through solenoid

L is length along which current pass = 0.350 m

plugging value to get required value of current

1.00*10^{-4} = \frac{1140*4*\pi *10^{-7} I}{0.350}

I = 2.4 *10^{-2} A = 2.4 mA

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RideAnS [48]

Answer:

C=2.32\times 10^{-4}\ Ns^2/m^2

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, F_r=Cv^2

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

Cv^2=mg

C=\dfrac{mg}{v^2}

C=\dfrac{0.046\times 9.8}{(44)^2}

C=2.32\times 10^{-4}\ Ns^2/m^2

Hence, this is the required solution.

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Which statements about Earth’s core help explain Earth’s magnetic field? Check all that apply.
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A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b
Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = \frac{1}{2} k x^{2}

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = \frac{1}{2} k x^{2}

10 = \frac{1}{2} k 10^{2}

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = \frac{1}{2} k x^{2}

20 = \frac{1}{2} (0.2) x^{2}

40 = 0.2 x^{2}

x^{2} = 200

x = \sqrt{200}

  = 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

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A rocket moves upward from rest with an acceleration of 40 m/s2 for 5 seconds. It then runs out of fuel and continues to move up
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Answer:

Maximum height of rocket  = 2538.74 m

Explanation:

We have equation of motion s = ut + 0.5 at²

For first 5 seconds

          s = 0 x 5 + 0.5 x 40 x 5² = 500 m

Now let us find out time after 5 seconds rocket move upward.

We have the equation of motion v = u + at

After 5 seconds velocity of rocket

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After 5 seconds the velocity reduces 9.8m/s per second due to gravity.

Time of flying after 5 seconds

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Distance traveled in this 20.38 s

          s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m

Maximum height of rocket = 500 +2038.74 = 2538.74 m

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