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3 years ago

When a battery does 24 j of work on 10 c of charge, the voltage it supplies is?

2 answers:

7 0

W = 24 J ( W = work done)
C = 10C (C = charge)
therefore,
V(voltage) = W/C

V = 24/10
V = 2.4V
therefore
the voltage supplied is 2.4 Volts
 or 2.4 V

Readme [11.4K]3 years ago

6 0

1 volt = 1 Joule per Coulomb.

24 J / 10 C = 2.4 J/C = 2.4 volts

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In Which figure below is the trend line drawn correctly

saul85 [17]

Answer:

The first graph is the trend line.

Explanation:

Trend Line

It's a line indicating the tendency of something being measured or evaluated. The trend line is meant to represent an approximation of a set of points of a measured magnitude.

We have three possible 'trend lines' to choose from. The second is not a line, so it's not the correct option. The third is not a line either, it's just a graph who joins all the measured points. It doesn't mark a trend.

The first graph is a trend line because it's a single line and can be used to represent the whole set of values .

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3 years ago

Consider a portion of a cell membrane that has a thickness of 7.50nm and 1.3 micrometers x 1.3 micrometers in area. A measuremen

lorasvet [3.4K]

Answer:

(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.

(1) is correct option

Explanation:

Given that,

Thickness = 7.50 nm

Area $A=(1.3\times1.3\times10^{-6})^2$

Potential difference = 92.2 mV

Resistivity of the material $\rho=1.30\times10^{7}\ \ohm$

We need to calculate the resistance

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.30\times10^{7}\times7.50\times10^{-9}}{(1.3\times1.3\times10^{-6})^2}$

$R=3.413\times10^{10}\ \Omega$

(a). We need to calculate the amount of current that flows through this portion of the membrane

Using Ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{92.2\times10^{-3}}{3.413\times10^{10}}$

$I=2.70\times10^{-12}\ A$

The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). If the side dimensions of the membrane portion is halved

We need to calculate the new resistance

Using formula of resistivity

$R'=\dfrac{\rho \dfrac{l}{2}}{A}$

Put the value into the formula

$R'=\dfrac{1.30\times10^{7}\times\dfrac{7.50\times10^{-9}}{2}}{(1.3\times1.3\times10^{-6})^2}$

$R'=1.706\times10^{10}\ \Omega$

We need to calculate the new current

Using Ohm's law

$V=I'R'$

$I'=\dfrac{V}{R'}$

Put the value into the formula

$I'=\dfrac{92.2\times10^{-3}}{1.706\times10^{10}}$

$I'=5.404\times10^{-12}\ A$

We need to calculate the factor

$\dfrac{I'}{I}=\dfrac{5.404\times10^{-12}}{2.70\times10^{-12}}$

$I'=2I$

The factor of the current is increase by factor of 2.

Hence,(a). The amount of current that flows through this portion of the membrane is $2.70\times10^{-12}\ A$

(b). The factor of the current is increase by factor of 2.

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