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snow_tiger [21]
3 years ago
11

2. An ammeter registers 2.5 A of current in a wire that is connected to a 9.0 V

Physics
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

3.6 Ω

Explanation:

Resistance: This is defined as the opposition to the flow of current in an electric circuit. The S.I unit of Resistance is Ohm's (Ω).

From the question above,

Applying Ohm's law

V = IR...................... Equation 1

Where V = Voltage, I = current registered by the ammeter, R = Resistance.

make R the subject of the equation

R = V/I............... Equation 2

Given: V = 9 V, I = 2.5 A

Substitute into equation 2

R = 9/2.5

R = 3.6 Ω

Hence the resistance of the wire =  3.6 Ω

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Explanation:

Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let's suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the source charge. The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force - either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge on the test charge.

7 0
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What type of eclipse is shown? How do you know?
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8 0
4 years ago
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A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant
djverab [1.8K]

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference,\frac{dC}{dt}=12 cm/s

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We have to find the induced emf at time t=9 s

We know that induced amf,E=\frac{Bd(A)}{dt}

Area of circular coil,A=\pi r^2

E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}

Circumference of circular coil,C=2\pi r

165=2\pi r

r=\frac{165}{2\pi}

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s

Radius of coil at time t=9 s

r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m

1 m=100 cm

E=-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V

Magnitude of induced emf=0.005 V

4 0
3 years ago
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During an episode of turbulence in an airplane you feel 210 n heavier than usual.if your mass is 72 kg, what are the magnitude a
lana66690 [7]
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Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.
8 0
4 years ago
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