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What is the equivalent resistance of a circuit that contains four 75.0 resistors connected in parallel with a 100.0 V battery?

zhannawk [14.2K]

To get the total resistance in a parallel circuit, you need to remember that unlike in a series, you do not just merely add the resistances. You need to get the reciprocal first of each resistance and add them together.

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}... +\frac{1}{R_{n}}$

After adding them, you will get the reciprocal again and then compute for the value. The problem says that there are 4 resistors in the circuit that have a resistance of 75.

$\frac{1}{R_{T}} = \frac{1}{75}+\frac{1}{75}+\frac{1}{75}+\frac{1}{75}$

Add up the numerator and copy the denominator:

$\frac{1}{R_{T}} = \frac{4}{75}$

Then get the reciprocal to get the total resistance:

$R _{T} = \frac{75}{4} = 18.75$

The answer to your question then is A. 18.8.

5 0

3 years ago

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A cabbie is trying to stop when he notices a fare is whistling them over. The

liberstina [14]

K.E=18750J
Mass=m=2100kg
Velocity=v

$\boxed{\sf K.E=\dfrac{1}{2}mv^2}$

$\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2$

$\\ \sf\longmapsto 18750=1050v^2$

$\\ \sf\longmapsto v^2=\dfrac{18750}{1050}$

$\\ \sf\longmapsto v^2=17.85m^2$

$\\ \sf\longmapsto v=\sqrt{17.85}$

$\\ \sf\longmapsto v=4.1m/s$

7 0

3 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain

Lesechka [4]

Answer: $4.21 \times 10^{-10} J/cm^4$

$1 kPa= 10^3 Pa$

$1 km=10^5 cm$

$1kPa/km=0.01 Pa/cm$

$1kPa/km=10^{-8} J/cm^4$

$\Rightarrow r= 0.0421 kPa/km= 0.0421 kPa/km \times \frac{10^{-8} J/cm^4}{1 kPa/km}= 0.0421 \times 10^{-8}J/cm^4=4.21 \times 10^{-10} J/cm^4$

3 0

3 years ago

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suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>

6 0

3 years ago

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