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3 years ago

Select the correct answer.

Physics

2 answers:

Kryger [21]3 years ago

6 0

Answer:

8.36*10^-4N/C

Explanation:

To find the electric field you use the following formula:

$F=qE\\\\E=\frac{F}{q}$

F: electric force = 8.2*10^-2 N

q: charge of the balloon = 9.4*10 C

By replacing F and q you obtain:

$E=\frac{8.2*10^{-2}N}{9.4*10}=8.36*10^{-4}\frac{N}{C}$

hence, the electric field is 8.36*10^-4N/C

kvasek [131]3 years ago

4 0

Answer:

8.37×10⁻⁴ N/C

Explanation:

Electric Field: This is the ratio of electrostatic force to electric charge. The S.I unit of electric field is N/C.

From the question, the expression for electric field is given as,

E = F/Q.......................... Equation 1

Where E = Electric Field, F = force experienced by the charged balloon, Q = Charge on the balloon.

Given: F = 8.2×10⁻² Newton, Q = 9.8×10 Coulombs = 98 Coulombs

Substitute these values into equation 1

E = 8.2×10⁻² /98

E = 8.37×10⁻⁴ N/C

Hence the Electric Field of the charged balloon = 8.37×10⁻⁴ N/C

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A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

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Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

Answer:

The heat transferred is $Q = 5.866 * 10^9 J$

The power is $P = 5866\ MW$

Explanation:

From the question we are told that

Mass of the water per second is $m = 1917 \ kg$

The initial temperature of the water is $T_i = 35^oC$

The boiling point of water is $T_b = 100^oC$

The final temperature $T_f = 450^oC$

The latent heat of vapourization of water is $c__{L}} = 2256*10^3 J/kg$

The specific heat of water $c_w = 4184 J/kg^oC$

The specific heat of stem is $C_s =1520 \ J/kg ^oC$

Generally the heat needed each second is mathematically represented as

$Q = m[c_w (T_i - T_b) + m* c__{L}} + m* c__{S}} (T_f - T_b)]$

Then substituting the value

$Q = m[c_w [T_i - T_b] + c__{L}} + C__{S}} [T_f - T_b]]$

$Q = 1917 [(4184) [100 - 35] + [2256 * 10^3] +[1520] [450 - 100]]$

$Q = 1917 * [3.05996 * 10^6]$

$Q = 5.866 * 10^9 J$

The power required is mathematically represented as

$P = \frac{Q}{t}$

From the question $t = 1\ s$

$P = \frac{5.866 *10^9}{1}$

$P = 5866*10^6 \ W$

$P = 5866\ MW$

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Explanation:

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