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MissTica
3 years ago
6

) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tigh

tly wound around the solenoid at its center. The mutual inductance of the coil and solenoid is At a given instant, the current in the solenoid is and is decreasing at the rate of At the given instant, what is the induced current in the coil
Physics
1 answer:
Aleonysh [2.5K]3 years ago
5 0

Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil

Answer:

6.169 μA

Explanation:

Formula for induced EMF is given by the equation;

EMF = M(di/dt). We are given;

di/dt = 2.5 A/s

M = 19μH = 19 × 10^(-6) H

Thus;

EMF = 19 × 10^(-6) × 2.5.

EMF = 47.5 × 10^(-6) V

Formula for current is;

i = EMF/R. R is resistance given as 7.7 ohms.

Thus; i = 47.5 × 10^(-6)/7.7

i = 6.169 μA

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A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib
Natasha_Volkova [10]

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

L_1A_1 = L_2A_2

1.20(\pi 18^2) = L_2(\pi 5^2)

L_2 = 15.55 m

so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m

So here the net pressure on the smaller area is given as

P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)

excess pressure exerted on the smaller area is given as

P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)

P_{ex} = 2.49\times 10^5 Pascal

now the force required on the other side is given as

F = P_ex (area)

F = (2.49 \times 10^5)(\pi (0.05)^2)

F = 1958.4 N

3 0
3 years ago
A 2-kg box is pushed by a force of 4 N for 2 seconds. It has an initial velocity vo = 2 m/s to the right. NOTE: Since this probl
IRISSAK [1]

Answer:

Kf= 36 J

W(net) = 32 J

Explanation:

Given that

m = 2 kg

F= 4 N

t= 2 s

Initial velocity ,u= 2 m/s

We know that rate of change of linear momentum is called force.

F= dP/dt

F.t = ΔP

ΔP = Pf - Pi

ΔP = m v  - m u

v= Final velocity

By putting the values

4 x 2 = 2 ( v - 2)

8 =  2 ( v - 2)

4 = v - 2

v= 6 m/s

The final kinetic energy Kf

Kf= 1/2 m v²

Kf= 0.5 x 2 x 6²

Kf= 36 J

Initial kinetic energy Ki

Ki = 1/2 m u²

Ki= 0.5 x 2 x 2²

Ki = 4 J

We know that net work is equal to the change in kinetic energy

W(net) = Kf - Ki

W(net) = 36 - 4

W(net) = 32 J

7 0
3 years ago
Calculate the wavelength of an orange light wave with a frequency of 5.085 x 10^14 Hz. The speed of light is 3.0 x 10^8 m/s.
Margarita [4]

Answer:

5.9 x 10⁻⁷m

Explanation:

Given parameters:

Frequency = 5.085 x 10¹⁴Hz

Speed of light  = 3.0 x 10⁸m/s

Unknown:

Wavelength of the orange light  = ?

Solution:

The wavelength can be derived using the expression below;

            wavelength  = \frac{v}{f}

v is the speed of light

f is the frequency

            wavelength  = \frac{3 x 10^{8} }{5.085 x 10^{14} }   = 5.9 x 10⁻⁷m

3 0
3 years ago
Explain why astronomers use the term "blueshifted" for objects moving toward us and "redshifted" for objects moving away from us
ValentinkaMS [17]
<span>Two of them are "redshift" and "blueshift", which are used to describe an object'smotion toward or away from us in space. Redshift indicates that an object is moving away from us. "Blueshift" is a term that astronomers use to describe an object that is moving toward another object or toward us.</span>
8 0
3 years ago
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 x 108 m. Fin
gizmo_the_mogwai [7]

Answer:

a). 1.28333 seconds

b). 186.66 seconds

Explanation:

a). Given :

Distance between the earth and the moon, d = $3.85 \times 10^8$ m

Speed of the radio waves, c = $3 \times 10^8$ m/s

Therefore the time required for the voice of Neil Armstrong to reach the earth via radio waves is given by :

$t=\frac{d}{c}$

 $=\frac{3.85 \times 10^8}{3 \times 10^8}$

 = 1.28333 seconds

b). Distance between Mars and the earth, d = $5.6 \times 10^{10}$ m

   Speed of the radio waves, c = $3 \times 10^8$ m/s

So, the time required for his voice to reach earth is :

$t=\frac{d}{c}$

 $=\frac{5.6 \times 10^{10}}{3 \times 10^8}$

 = 186.66 seconds

6 0
3 years ago
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