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Irina-Kira [14]

4 years ago

8

What causes the electric charges to flow from one end of the battery to the other? a balance in electric potential a balance in

resistance a difference in electric potential a difference in resistance

2 answers:

Vilka [71]4 years ago

8 0

I think the correct answer from the choices listed above is the third option. It is the difference in electrical potential that causes the electric charges to flow from one end of the battery to the other. Hope this answers the question. Have a nice day.

andreev551 [17]4 years ago

7 0

Answer: Option (c) is the correct answer.

Explanation:

When a charge moves from a reference point to a specific point then the amount of work done per unit charge to move the charge is known as electric potential.

The difference in potential causes the charges to move from one point to another.

Thus, we can conclude that a difference in electric potential causes the electric charges to flow from one end of the battery to the other.

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The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

3 0

3 years ago

A car drives around a curve with a radius of 42 m at a velocity of 24 m/s. What is the centripetal acceleration of the car?

defon

Hope this helps you.

8 0

4 years ago

Read 2 more answers

Density is a property of matter that relates

tiny-mole [99]

Density is the mass of a substance per unit volume.

7 0

3 years ago

Aluminum has a density of

agasfer [191]

<h3><u>Volume is 0.1848 m³</u></h3><h3 />

Explanation:

<h2>Given:</h2>

m = 49.9 kg

ρ = 270 kg/m³

<h2>Required:</h2>

volume

<h2>Equation:</h2>

$ρ \:= \:\frac{m}{v}$

where: ρ - density

m - mass

v - volume

<h2>Solution:</h2>

Substitute the value of ρ and m

$ρ \:= \:\frac{m}{v}$

$270\: kg/m³\:= \:\frac{49.9\:kg}{v}$

$(v)\:270\: kg/m³\:= \:49.9\:kg$

$v\:= \:\frac{49.9\:kg}{270\: kg/m³}$

$v\:= \:0.1848\:m³$

<h2>Final Answer:</h2><h3><u>Volume is 0.1848 m³</u></h3>

3 0

3 years ago

Read 2 more answers

A boat has a mass of 4040 kg. Its engines generate a drive force of 4660 N due west, while the wind exerts a force of 880 N due

maxonik [38]

Answer:

Explanation:

Given:

Mass of the boat, m = 4040 kg

The driving force of engine, FB = 4660 N in west = + 4660 N

The force of wind, Fwi = 880 N in east = -880 N

The force of water, Fwa = 1400 N in east = -1400N

Total three forces are acting on the boat

Fnet= Fb+fwi+Fwa

Fnet= 4660 - 880 - 1400

Fnet= +2380N

Acceleration (a) = Force/mass

= 2380/4040

= 0.59m/s2

6 0

3 years ago

Read 2 more answers