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2 years ago

Current in a conductor can be increased by:

Physics

2 answers:

larisa [96]2 years ago

8 0

THE CURRENT IN A CONDUCTOR CAN BE INCREASED BY BOTH INCREASING THE VOLTAGE ACROSS IT AND DECREASING ITS RESISTANCE.

Explanation:

Ohm's law states that the voltage or potential difference between two points is directly proportional to current passing through resistance and directly proportional to resistance.

According to ohm's law,

V = I * R

It means voltage is directly proportional to current and resistance.

current and resistance are inversely proportional to each other.

so from the above law it can be stated that current in a conductor can be increased by both increasing the voltage across it and decreasing its resistance.

Assoli18 [71]2 years ago

7 0

Current in a conductor can be increased by Both of these (3rd Option )

<u>Explanation:</u>

Ohm's law states that the current is directly proportional to its applied voltage and it is inversely proportional to the resistance between the points.

Factors increasing the current in a conductor are:

As we know that resistance and conductance are inversely proportional to each other.

As we increase the voltage across the two points, then the current in the conductor increases.

As we decrease the resistance the current in the conductor increases.

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3 years ago

If you hold a 50 kilogram barbell above your head for 3 seconds and

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Answer:

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3 years ago

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3 years ago

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Amanda [17]

Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

<h3>How many cells are produced? </h3>

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6 0

2 years ago

A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu

kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) = ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i) work required to pump the contents to the rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the cone's rim

Extra work required = 28,914.229 lb × 5ft = 144571.145 lb.ft

Total work required = (281,913.733 + 144571.145) lb.ft

= 426,484.878 lb.ft

5 0

2 years ago