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Travka [436]
3 years ago
6

a DJ uses a 110-volt outlet to plug in a strobe light. if the current flowing through the light is 0.050 amps, how much resistan

ce is within the circuit?
Physics
1 answer:
tigry1 [53]3 years ago
5 0

Ohms Law: V = IR

V is the voltage in volts

I is the current in amps

R is the resistance in Ohms

Rearrange: R = V/I

R = (110)/(0.050)

R = 2200

There are 2200 Ohms of resistance in the circuit.

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Explanation:

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At the beginning of a basketball game, the referee tosses the ball straight up with a speed of 4.6m/s. A player cannot touch the
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Answer:

t=0.47s

Explanation:

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g=gravity=-9.8m/s^2

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T=(Vf-Vo)/g

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2 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

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The blank distance is your answer

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