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3 years ago

Does gravity have an effect on the potential difference of a battery?

Physics

1 answer:

Sonja [21]3 years ago

5 0

Answer: The correct answer is "No".

Explanation:

Gravity: It is the force which causes object to fall on the earth. It is the force which attracts bodies towards each other.

Potential difference: It is defined as the potential acting between the two points. The work done in moving the unit positive charge from one location to the another location.

The potential difference in battery is caused by the electrodes. There are two terminals in battery: Negative terminal which is at lower potential and Positive terminal which is at higher potential. It forces the electrons to flow in the circuit which constitutes the current.

The gravity and the potential difference have no relation between them.

Therefore, gravity have no effect on the potential difference of a battery.

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3 years ago

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p

Temka [501]

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT............................... Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

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3 years ago

A ball is dropped from rest from a high window of a tall building and falls for 4 seconds. Neglecting air resistance, the final

Semenov [28]

Answer:

The final velocity of the ball is 39.2 m/s.

Explanation:

Given that,

A ball is dropped from rest from a high window of a tall building.

Time = 4 sec

We need to calculate the final velocity of the ball

Using equation if motion

$v=u+gt$

Where, v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Put the value into the formula

$v=0+9.8\times4$

$v=39.2\ m/s$

Hence, The final velocity of the ball is 39.2 m/s.

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3 years ago

A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I

natulia [17]

Answer:

x = 25 / μ [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

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N_B + N_A - W_van - W_load = 0

N_B + N_A = W_van + W_load

X axis

fr = ma

a = fr / m

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m = (W_van + W_load) / g

the friction force has the expression

fr = μ N_{total}

fr = μy (W_van + W_load)

we substitute

a = μ (W_van + W_load) $\frac{g}{W_van + W_load}$

a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

v² = v₀² - 2 a x

0 = v₀² -2a x

x = $\frac{v_o^2}{2a}$

x = $\frac{v_o^2}{2 \mu g}$

x = $\frac{40^2}{2 \ 32 \ \mu}$

x = 25 / μ [ ft]

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2 years ago