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1 month ago

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A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o

f the surface of this sphere and find it to be 14 kV relative to infinity. Find the excess charge on this sphere.

1 answer:

Maru [3.3K]1 month ago

7 0

Answer:

The excess charge is $Q = 3.5 *10^{-7} \ C$

Explanation:

According to the question, we are informed that

The diameter is $d = 45 \ cm = 0.45 \ m$

The potential of the surface is $V = 14 \ kV = 14 *10^{3} \ V$

The radius of the sphere is

$r = \frac{d}{2}$

by plugging in given values

$r = \frac{0.45}{2}$

$r = 0.225 \ m$

The potential at the surface is mathematically expressed as

$V = \frac{k * Q }{r }$

Where k is Coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

Based on the question stating there are no other charges, Q represents the excess charge

Therefore

$Q = \frac{V* r}{ k}$

inserting the numerical values

$Q = \frac{14 *10^{3} 0.225}{ 9*10^9}$

$Q = 3.5 *10^{-7} \ C$

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Answer:

C) True. The distance S increases over time, with v₁ = gt and v₂ = g (t-t₀), illustrating that v₁> v₂ for the same t.

Explanation:

We have a set of statements to evaluate for correctness. The most effective approach is to examine the problem in detail.

Using the equation for vertical launch, we acknowledge that the positive direction signifies downward movement.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Stone 2

Released shortly thereafter, let's assume a delay of one second, we can utilize the same timing mechanism

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

We can now calculate the separation distance between the two stones, which is applicable for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t² - 2 t t₀ + t₀²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t - t₀)

This represents the distance between the two stones over time, with the coefficient outside the parentheses being constant.

For t < to, the first stone remains stationary while the distance grows.

For t > = to, the expression (2t/to-1) yields a value greater than 1, indicating that the distance expands as time progresses.

We can now analyze the different statements

A) false. The height difference increases over time.

B) False S increases.

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3 months ago

A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be

Yuliya22 [3333]

Answer:

The work done, W = 19.6 J

Explanation:

It’s provided that

The mass of the block, m = 5 kg

The velocity of the block, v = 10 m/s

The coefficient of kinetic friction between the block and rough surface is 0.2

Distance traveled by the block, d = 2 m

As the block traverses the rough section, it loses energy equal to the work done by the kinetic energy.

$W=\mu_kmgd$

$W=0.2\times 5\times 9.8\times 2$

W = 19.6 J

Thus, the change in kinetic energy of the block moving through the rough section is 19.6 J. Consequently, this is the required answer.

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2 months ago

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

Ostrovityanka [3204]

Response:

$\text{heat loss} = 24864.05 \ W/m^2$

Clarification:

If

$T_1$ , $T_2$ represent the temperatures of gases and liquids in Kelvins,

$t_1$ and $t_2$ denote the thicknesses of the gas layer and steel slab in meters,
$h_1$ , $h_2$ are the convection coefficients for gas and liquid in $W/m^2 \cdot K$ ,
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and $k_1, k_2$ signify thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

by employing known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Utilizing the rate equation:

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature is $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Correspondingly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature profile is depicted in the image provided

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The work done can be calculated using the equation:

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Where:

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Vo denotes the initial speed = 72 kph = 20 m/s

By substituting in the given values:

Work = (1/2)*2500*(25^2) - (1/2)*2500*(20^2) = 281250 J, which can also be represented as 2.8 x 10^5 Joules.

The correct choice among the options is A.

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3 months ago

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