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3 years ago

Why are carbon brushes used? How do these work?

Physics

2 answers:

andreev551 [17]3 years ago

7 0

Answer:

A carbon brush is a sliding contact used to transmit electrical current from a static to a rotating part in a motor or generator, and machines, ensuring a spark free communication. Carbon brushes are also used in electric motors to provide a path for flow of current.

Explanation:

Zepler [3.9K]3 years ago

6 0

Answer:

Why are they used:

A carbon brush is a sliding contact used to transmit electrical current from a static to a rotating part in a motor or generator, and, as regards DC machines, ensuring a spark-free commutation.

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Which of the following is not a rock? a. granite c. marble b. shale d. diamond

bekas [8.4K]

The correct answer is D, Diamond

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3 years ago

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A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r

elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

$V_1=26.8 L$

$P_1=744mm Hg$

$T_1=31.2^{\circ} C=31.2+273=304.2K$

$P_2=385mmHg$

$T_2=-14.8^{\circ}=273-14.8=258.2K$

We know that

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Substitute the values

$V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}$

$V_2=43.96 L$

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

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3 years ago

what is the cost to run a single 100 w light bulb for 30 days if the electric company charges $0.18 for every KW-hr?

xxTIMURxx [149]

Explanation:

power of the bulb = 100w=0.1kw

power consumption when the bulb is used for 30 days ( 30 *24 = 720 hours) =0.1*720=72kwh

cost to run the bulb for 30 days = 72*0.18

= $12.96

Answer is $12.96..................

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3 years ago

A 1.51 kg ball and a 1.97 kg ball are connected by a 1.63 m long rigid, massless rod. The rod is rotating clockwise about its ce

Eddi Din [679]

Answer:

$T = 1.205\,N\cdot m$

Explanation:

Needed torque can be estimated by means of the Theorem of Angular Momentum Conservation and Impact Theorem. The center of mass of the system is:

$\bar r = \frac{(0\,m)\cdot (1.51\,kg)+(1.63\,kg)\cdot (1.97\,kg)}{1.51\,kg+1.97\,kg}$

$\bar r = 0.923\,m$

Let assume that both masses can be modelled as particles, then:

$[(1.51\,kg)\cdot (0.923\,m)^{2} + (1.97\,kg)\cdot (0.707\,m)^{2}]\cdot (38\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} ) -T\cdot (7.5\,s) = 0\,\frac{kg\cdot m^{2}}{s}$

The torque needed to stop the system is:

$T = 1.205\,N\cdot m$

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3 years ago

How much does the earth weigh

tatiyna

Answer:

5.972 × 10^24 kg

Explanation:

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3 years ago

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