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disa [49]
3 years ago
14

What resistance would produce a current of 200A with a potiential difference of 200V?

Physics
1 answer:
viktelen [127]3 years ago
5 0
Aw, I hate physics, is this on Apex?

Resistance can be calculated with the information given in the question.
Equation for Resistance: R = V/I
V (voltage) = 200 Volts
I (current) = 200 Amps

So 200 divided by 200 = freaking 1

Answer: R = 1 (ohms)

Hope this Helps!

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A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c
Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}

hence, the speed of the spheres is 37.45 m/s

8 0
3 years ago
True or False: <br><br> Solids always<br> have a higher density than<br> liquids and gases.
Hunter-Best [27]

Answer:

<h3>False</h3>

<h3> I hoped this helpful for you....</h3>

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7 0
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The heater element of a 120 V toaster is a 5.4 m length of nichrome wire, whose diameter is 0.48 mm. The resistivity of nichrome
Ksenya-84 [330]

Answer:

The power drawn by the toaster is closest to:

(A) 370 W

Explanation:

First we calculate the resistance of the nichrome wire (R).

R=\frac{pL}{A} =\frac{pL}{\pi r^{2} } \\

Where radious (r), resistance coefficient (p), and Length (L)

r=\frac{0.48}{2} 10^{-3} m\\\\L=5.4 m\\p=1.3*10^{-3}\varOmega  \\\\\\Replacing:\\\\R=\frac{1.3(10)^{-6} *5.4}{\pi* 0.24^{2} (10)^{-6}} =38.7940

After replace the value in the ohm law power formula to obtain the power consumed:

P=\frac{V^2}{R} =\frac{120^2}{38.7940} =371.191 Watts

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Answer:

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Explanation:

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