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scoundrel [369]
2 years ago
14

A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads

. Find the amount of electric charge that resides on each capacitor once it is fully charged.

Physics
1 answer:
Bond [772]2 years ago
3 0

The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is calculated as follows;

Capacitors in series;

1/Ct = 1/8 + 1/7.5

1/Ct = 0.25833

Ct = 3.87 mF

Capacitors is parallel;

Ct = 3.87 mF + 12 mF + 15 mF

Ct = 30.87 mF

Ct = 0.03087 F

<h3>Charge in each capacitor</h3>

Q = CV

Q = 0.03087 x 12

Q = 0.37 C

Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.

Learn more about capacitors here:  brainly.com/question/13578522

#SPJ1

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After you enlarge a map, which one of the following scale remains correct?
9966 [12]

Answer:

None

Explanation:

An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:

  • Verbal scale use of words to represent scale information on the map.  The distance or linear units are used for depicting this scale on the map.  For example: 1 inch = 1 Kilo meter.
  • Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio.  Example: 1: 10,000 or 1/10,000
  • In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given.  Example: 1:500
  • Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map.  Example: 1:25,000
  • A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.

I hope you find this information useful and interesting! Good luck!

6 0
3 years ago
What is the displacement of the particle in the time interval 7 seconds to 8 seconds? A. 0 meters B. 1.5 meters C. 3 meters D. 7
Dmitrij [34]

Answer: B. 1.5 meters

Explanation:

4 0
3 years ago
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After asking a question, a scientist can form a(n) , which is an idea that may be proved or disproved by an experiment.
DochEvi [55]

Answer:

When scientists have a question, they form a hypothesis, <em>which</em><em> </em><em>is</em><em> </em><em>an</em><em> </em><em>idea</em><em> </em><em>that</em><em> </em><em>may</em><em> </em><em>be</em><em> </em><em>proved</em><em> </em><em>or</em><em> </em><em>disproved</em><em> </em><em>by</em><em> </em><em>an</em><em> </em><em>experiment</em><em>.</em>

6 0
3 years ago
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*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co
sp2606 [1]

Answer: The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

I = \sqrt{\frac{P}{R} }

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }

Then, simplify the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} }

Rationalize the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}

Simplify the numerator by finding its factors:

I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}

The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

8 0
3 years ago
A spacecraft and a staellite are at diametrically opposite position in the same circular orbit of altitude 500 km above the eart
Tanya [424]

Answer:

Hello the diagram related to your question is attached below

answer: a) 851 m/s

             b)  8506.1 secs

Explanation:

calculate the periodic time of the satellite using the equation below

t = \frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }  --  ( 1 )

where ; R = 6370 km

h = 500 km

g = 9.81 m/s^2

input given values into equation 1

t = 5670.75 secs

next calculate the periodic time taken by the space craft  

<u>a) determine the increase in speed </u>

V = v - \sqrt{\frac{gR^2}{R + h} }  

where ; v = 8463 m/s , R = 6370 km, h = 500 km

V = 851 m/s

b) Determine the periodic time for the elliptic orbit

τ = \frac{3t}{2}

 = \frac{3*5670.76}{2}  =  8506.1 secs

attached below is the remaining part of the detailed solution

5 0
3 years ago
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