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GREYUIT [131]
3 years ago
9

The potential of electrons in a circuit can be increased by _____. flowing through the conducting wires the action of a battery

passing through a circuit element water pressure
Physics
2 answers:
Zarrin [17]3 years ago
7 0

Answer:

The potential of electrons in a circuit can be increased by the action of the battery.

Explanation:

The potential of electrons in a circuit can be increased by the action of the battery. This is because the battery provides a large amount of power that results in a strong voltage propulsion that is able to increase the potential of elestrons in a circuit. An example of this is the use of a battery in an electric bicycle. The bike only works by supplying battery power through a voltage that increases the electron circuit creating the electric power.

Kaylis [27]3 years ago
6 0
The potential of electrons in a circuit can be increased by the action of a battery.
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Relative means
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Answer:As a noun relative means a person who is connected with others by blood.

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2 years ago
Assume you need to design a hydronic system that can deliver 80,000 Btu/hr. What flow rate of water is required if the temperatu
PolarNik [594]

Answer:

At 10°F change in temperature

Mass flowrate = 1.01 kg/s = 2.227 lbm/s

Volumetric flowrate = 1010 m³/s = 35667.8 ft³/s

At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

ṁ = 23445.7/(4182 × 11.111)

ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

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3 years ago
Using the image provided, which moon phase is next?
ella [17]

Answer:

Waning Crescent

5 0
2 years ago
Read 2 more answers
3 facts about the concept or term of tectonic plates 
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1) Plates movement happens very slowly.
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8 0
3 years ago
Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o
KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹  N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹  N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea =  +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹  N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb =  -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴  +  -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0
3 years ago
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