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klasskru [66]
4 years ago
13

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.160-T magnetic field near the center o

f the solenoid. You have enough wire for 4300 circular turns. This solenoid must be 1.10 m long and 3.50 cm in diameter. What current will you need to produce the necessary field?
Physics
1 answer:
MariettaO [177]4 years ago
5 0

Answer:

Current = 32.6 A

Explanation:

Magnetic field = B = 0.160 T

No. of turns of coil = N = 4300

Length of the solenoid = L = 1.10 m

Current required depends on the B, L , N and the magnetic permeability μ₀

B =  μ₀ N I / L

⇒ Current =  I = B L /  μ₀N = ( 0.160 ) ( 1.10) / (4π × 10⁻⁷)(4300)

                                           =  32.6 A

   

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Which elements are not likely to be formed in the sun sometime during its life cycle?
Sladkaya [172]

Answer:

correct answer is iron

Explanation:

solution

as Iron is only formed in the cores of the star that is more massive than the sun because it takes very much energy  as gravitational force push inward on the star  

and the sun can form the helium and  carbon and the oxygen and also other elements  

but not the iron or any iron group element

so correct answer is iron

8 0
3 years ago
The second law of thermodynamics states that
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3 years ago
How would we be able to detect a large asteroid if it were heading straight for earth?
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4 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
2 years ago
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allochka39001 [22]

The gravitational force between them is <em>1.25 x 10^20 Newtons.</em>

(I think you must have typed the mass of the moon wrong.  

It must be 6.0 x 10^22 kg.)

4 0
4 years ago
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