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larisa86 [58]
3 years ago
12

A superconducting solenoid has 3300 turns per meter and can carry a maximum current of 4.1 ka. find the magnetic field strength

in the solenoid.

Physics
2 answers:
jeyben [28]3 years ago
8 0
B = μ x n x I = 1.25 x 10^-6 x 3300 x 4.1 x 10^3
                    =  16.91 T
avanturin [10]3 years ago
7 0
Hope this helps you.

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According to the first rule, if a force pulls on one end of a rope, the tension in the rope equals the magnitude of the pulling
mixer [17]

Answer:

F.

Explanation:

Here in the question the mass of the pulley is zero, hence, the tension in the cable throughout is same.

magnitude of tension in rope 1 is

T1= F

Hence the tension T1 is rope 1 is F.

5 0
3 years ago
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Use the concepts of kinetic energy and potential energy to describe the motion of a child on a swing. Why does the child need a
andrew-mc [135]
When the child is moving, he/she has kinetic energy. For just a brief second before they move the other way, the child is not moving, but they have gravitational potential energy.

The child may need a push from time to time because friction with the air causes loss of energy.
4 0
3 years ago
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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb
balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = F_1=5.6 lb

Area of the piston = A_1=0.6 inches^2

Force applied on the brakes = F_2

Area of the brakes = A_2=1.8 inches^2

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

\frac{F_1}{A_2}=\frac{F_2}{A_2}

F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb

16.8 lb is the force on the brake pad of one wheel.

5 0
3 years ago
Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1
Artemon [7]

Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

4 0
3 years ago
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