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3 years ago

Every complete circuit includes a device that provides emf. What type of quantity?

Physics

1 answer:

laiz [17]3 years ago

6 0

Answer:

energy per unit charge

Explanation:

EMF is energy per unit charge and has unit joule/ coulomb, where joule is unit of energy and coulomb is the unit of charge.

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. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content o

mote1985 [20]

Answer:

40479.6 J

Explanation:

Applying,

q = cm(t₂-t₁).................... Equation 1

Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C

Constant: c = 4200 J/kg.°C

Substitute these values into equation 1

q = 4200(0.079)(143-21)

q = 331.8(122)

q = 40479.6 J

4 0

3 years ago

Which of the following best describes the use of a renewable resource? (2 points) Most power plants burn fossil fuels to generat

kolbaska11 [484]

Which of the following best describes the use of a renewable resource?

Answer:

There are areas in California that use the heat from deep in the Earth to generate electricity.

Explanation:

It comes from the earth itself and we use a lot of things that comes from the earth and deep within it.

5 0

3 years ago

What is the moon phase today

Elis [28]

Waxxing Gibbous Phase

4 0

3 years ago

Read 2 more answers

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

Phosphate never enters the<br> O atmosphere<br> O ground<br> O water<br> O ocean

gizmo_the_mogwai [7]

I think you can google this because I really don’t know the answer I’m so sorry

7 0

3 years ago