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3 years ago

*Please Help!

1 answer:

8 0

We calculate by using the equation as follows:

Free Energy = DeltaG

The formula is DeltaG = Enthalpy - (Temperature)*(Entropy) 

DeltaG = 1.25 * 10^5 J - (290)(300) 

DeltaG or Free Energy = 38000 J/mol or 38kJ/mol. 

This means that this system is not spontaneous at 290 K

Hope this answers the question. Have a nice day.

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Autorhythmic cells can generate action potentials spontaneously because they have Autorhythmic cells can generate action potenti

OlgaM077 [116]

Answer:

They can generate potentials spontaneously because they have Unstable Membrane Potentials.

Explanation:

Autorythmic cells or Pacemaker cells are cells that provide Action potentials (electrical impulses) that starts off the cardiac cycle.

N:B This action potential is created spontaneously.

To explain further, the heart originate in specialized cardiac muscle cells, called autorhythmic cells, that can excite themselves and therefore are able to generate an action potential without external stimulation by nerve cells. And this sets the cardiac cycle i

(Pumping of the heart) into motion. (The pace maker potential)

The Autorhythmic cells create an action potential spontaneously

And this is possible because they have an UNSTABLE RESTING POTENTIAL that is continuously depolarizing, while it drifts slowly toward threshold. As Na+ ions enter the cell, the inner surface of the plasma membrane becomes less negative gradually, thus generating the pacemaker potential.

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3 years ago

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What is the net force on a car with a mass of 1000 kg if its acceleration is 35 m/s^2?

VashaNatasha [74]

Answer:

3000N

Explanation:

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2 years ago

b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure with dept

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3 years ago

two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?

Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force $F$ between two charged objects:

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}}$ ,

where

$k$ is coulomb's constant. $k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2}$ in vacuum.
$q_1$ and $q_2$ are the signed charge of the objects.
$r$ is the distance between the two objects.

For the two electrons:

$q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}$ .
$r = 1\times 10^{-10}\;\text{m}$ .

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}$ .

The sign of $F$ is negative. In other words, the two electrons repel each other since the signs of their charges are the same.

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3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0

3 years ago