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asambeis [7]
2 years ago
13

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

s h = 0.60 m, what is the current (in A) flowing through the rectangular toroid when the energy in its magnetic field is 4.0 ✕ 10−6 J?
Physics
1 answer:
andrey2020 [161]2 years ago
3 0

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field U=4\times 10^{-6}J

Value of inductance L =0.060 H

Energy stored in magnetic field is given by U=\frac{1}{2}Li^2

i=\sqrt{\frac{2U}{L}}

i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A

So the current flowing through rectangular toroid will be 0.01154 A

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Calculate the distance the marble travels during the first 3.0 seconds. [Show all work, including the equation and substitution
Alenkinab [10]

D = V0t + 0.5at^2

Where d is the distance

V0 is the initial velocity

A is the acceleration

T is time

From the graph a = 4/3 m/s2

D = 0(3) + 0.5( 4/3 m/s2) ( 3 s)^2

D = 6 m

3 0
1 year ago
The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass
Triss [41]

Answer:

\Delta x=245\ mm

Explanation:

Given:

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<u>Now, the force due to the mass:</u>

F=m.g

F=40\times 9.8

F=392\ N

<u>Compression in Spring:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{392}{1600}

\Delta x=0.245\ m

or

\Delta x=245\ mm

8 0
2 years ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

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B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car
Novay_Z [31]

Answer:

4.5 *10^{3}\ kgm/s^{2}

Explanation:

Given:

Mass= 1.00 * 10^{3} \ kilograms,\\Acceleration =4.5\  meters/second^{2}

As we know that

Force=Mass * Acceleration

Putting the value of mass and Acceleration  we get ,

Force\  = \ 1.00 * 10^{3} * 4.5\ m/s^{2}

Force\  =4.5 * \ 10^{3} \  kgm/s^{2}

Therefore Net force is :4.5 *10^{3}\ kgm/s^{2}

0 0
1 year ago
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