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3 years ago

The electric potential a distance r from a small charge is proportional to what power of the distance from the charge?

Physics

1 answer:

Gekata [30.6K]3 years ago

5 0

Answer:

The power of the distance is -1.

Explanation:

The equation for the electric potential of a point charge is given by $V= k\frac{q}{r}$

where V is the electric potential, k is Coulomb's constant (it has a value of $9.0*10^{9}$ with units $\frac{Nm^{2} }{c^{2} }$ ), q is the electric charge of the small charge and r is the distance from the charge.

Now, the power of a number is how many times we multiply that number by itself; we see r appears only once in the equation. So we know the power is 1. But we can see in the equation that k and q are divided by r, which means r is the denominator. This means the power of r is negative (-).

Therefore, the power of r is -1.

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Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

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Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

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A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temp

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Answer:

19.34°C

Explanation:

When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e

Q₁ = -Q₂ ----------------------(i)

{A} Q₁ is the heat gained by the ice and it is given by the sum of ;

(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]

Where;

m₁ = mass of ice = 0.0550kg

c₁ = a constant called specific heat capacity of ice = 2108J/kg°C

ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C

(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]

Where;

m₁ = mass of ice = 0.0550kg

L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg

Therefore,

Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁] ------------------(ii)

Substitute the values of m₁, c₁, ΔT₁ and L₁ into equation (ii) as follows;

Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]

Q₁ = [3478.2] + [18370]

Q₁ = 21848.2 J

{B} Q₂ is the heat lost by the hot water and is given by

Q₂ = m₂ x c₂ x ΔT₂ -----------------(iii)

Where;

m₂ = mass of water = 0.400kg

c₂ = a constant called specific heat capacity of water = 4200J/Kg°C

ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C

Substitute these values into equation (iii) as follows;

Q₂ = 0.400 x 4200 x (T - 35)

Q₂ = 1680 x (T-35) J

{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;

Q₁ = -Q₂

=> 21848.2 = - 1680 x (T-35)

=> 35 - T = 21848.2 / 1680

=> 35 - T = 13

=> T = 35 - 13

=> T = 22

Therefore the final temperature of the hot water is 22°C.

Now let's find the final temperature of the mixture.

The mixture contains hot water at 22°C and melted ice at 0°C

At this temperature, the heat ( $Q_{W}$ ) due to the hot water will be equal to the negative of the one ( $Q_{I}$ ) due to the melted ice.

i.e

$Q_{W}$ = - $Q_{I}$ -----------------(a)

Where;

$Q_{I}$ = $m_{I}$ x $c_{I}$ x Δ $T_{I}$ [ $m_{I}$ = mass of ice, $c_{I}$ = specific heat capacity of melted ice which is now water and Δ $T_{I}$ = change in temperature of the melted ice]

and

$Q_{W}$ = $m_{W}$ x $c_{W}$ x Δ $T_{W}$

[ $m_{W}$ = mass of water, $c_{W}$ = specific heat capacity of water and Δ $T_{W}$ = change in temperature of the water]

Substitute the values of $Q_{W}$ and $Q_{I}$ into equation (a) as follows

$m_{W}$ x $c_{W}$ x Δ $T_{W}$ = - $m_{I}$ x $c_{I}$ x Δ $T_{I}$

Note that $c_{W}$ and $c_{I}$ are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$ -----------------------(b)

Now, let's analyse Δ $T_{W}$ and Δ $T_{I}$ . The final temperature ( $T_{F}$ ) of the two kinds of water(melted ice and cooled water) are now the same.

=> Δ $T_{W}$ = change in temperature of water = final temperature of water( $T_{F}$ ) - initial temperature of water( $T_{IW}$ )

Δ $T_{W}$ = $T_{F}$ - $T_{IW}$

Where;

$T_{IW}$ = 22°C [which is the final temperature of water before mixture]

=> Δ $T_{I}$ = change in temperature of melted ice = final temperature of water( $T_{F}$ ) - initial temperature of melted ice ( $T_{II}$ )

Δ $T_{I}$ = $T_{F}$ - $T_{II}$

$T_{II}$ = 0°C (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$

0.400 x ( $T_{F}$ - $T_{IW}$ ) = -0.0550 x ( $T_{F}$ - $T_{II}$ )

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ - 0)

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ )

0.400 $T_{F}$ - 8.8 = -0.0550 $T_{F}$

0.400 $T_{F}$ + 0.0550 $T_{F}$ = 8.8

0.455 $T_{F}$ = 8.8

$T_{F}$ = 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

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