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mote1985 [20]
3 years ago
5

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th

at unrealistically large currents are needed to produce noticeable torques. Suppose a 20-cm-diameter loop of wire is oriented for maximum torque in the earth's field.What current would it need to carry in order to experience a very modest 1.0?
Physics
1 answer:
fiasKO [112]3 years ago
8 0

Answer:

636.619772368 A

Explanation:

\tau = Torque = 1\times 10^{-3}\ N/m

B = Magnetic field of Earth = 5\times 10^{-5}\ T

A = Area

d = Diameter = 20 cm

Current is given by

I=\dfrac{\tau}{BA}\\\Rightarrow I=\dfrac{1\times 10^{-3}}{5\times 10^{-5}\times \dfrac{\pi}{4}\times 0.2^2}\\\Rightarrow I=636.619772368\ A

The current is 636.619772368 A

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alukav5142 [94]
As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second

44.0 m/s - 2.0 m/s = 42.0 m/s 

42.0 m/s / 12 s = 3.5 m/s2

the acceleration of the rock would be 3.5 m/s2
4 0
3 years ago
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What force is left out of the Quantum Mechanics theory?
nirvana33 [79]

Answer:

Quantum mechanics is a key hypothesis in material science that gives a portrayal of the actual properties of nature at the size of iotas and subatomic particles. It is the establishment of all quantum physical science including quantum science, quantum field hypothesis, quantum innovation, and quantum data science.

Explanation:

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4 0
3 years ago
Um aluno pretende construir um aquecedor usando um enrolamento de fio. Experimenta e verifica que não
Vlad1618 [11]

Answer:

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6 0
3 years ago
How much force is required to accelerate 3 kg at 3 m/sec^2
bogdanovich [222]

Answer:

F=?

a=3m/s2

m=3kg

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Explanation:

7 0
3 years ago
Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm
Ghella [55]

Answer:

Total electric potential, V=1.32\times 10^6\ volts

Explanation:

It is given that,

First charge, q_1=3\ \mu C=3\times 10^{-6}\ C

Second charge, q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C

Distance of first charge from origin, r_1=1.15\ cm=0.0115\ m

Distance of second charge from origin, r_2=2\ cm=0.02\ m

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}

V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})

V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})

V = 1321826.08 V

or

V=1.32\times 10^6\ volts

So, the total electric potential at the origin is 1.32\times 10^6\ volts. Hence, this is the required solution.

3 0
3 years ago
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