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kari74 [83]
3 years ago
8

A circuit has a current of 1.2 A. If the voltage decreases to one-third of its original amount while the resistance remains cons

tant, what will be the resulting current?
Physics
1 answer:
lapo4ka [179]3 years ago
6 0
Consider the formula ...

Current  =  (voltage) / (resistance) .

If the resistance doesn't change, then the current is
directly proportional to the voltage.

If the voltage decreases to 1/3 of what it used to be,
then the current does the same thing.

1.2 A  ---->  0.4 A
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A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3
Klio2033 [76]

Answer:

\theta=30^{\circ}

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

F=ILB\ sin\theta

sin\theta=\dfrac{F}{ILB}

sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}

\theta=30^{\circ}

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

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A soccer ball is sitting in the middle of a soccer field during a game. When the referee blows his whistle, one of the players r
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How are pupae, larvae, and nymphs similar? *Answer Fast!*
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Answer:

They are all a cycle!

Explanation:

They just are all cycles.

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Which changes in an electric motor will make the motor stronger? Check all that apply. using a stronger permanent magnet using a
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A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

6 0
3 years ago
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