Ask question

Ask question

All categories

3 years ago

5

The room contains a refrigerator, operated by an electric motor. The motor does work at the rate of 438 W when it is running. Th

e refrigerator removes heat from the food storage space at a rate of 450 W when the motor is running. In an effort to cool the room, you open the refrigerator door and let the motor run continuously. At what net rate is heat added to (+) or subtracted from (−) the room and all of its contents?

1 answer:

yKpoI14uk [10]3 years ago

8 0

Answer: Heat is removed from the room at 12W

Explanation:

Energy (heat) is generated by the the rotor at a rate

∆Q/t = 438W

Work is done (heat removal) by the refrigerator at a rate

∆W/t = 450W

Net rate of heat removed =

∆Q/t - ∆W/t

= (438 - 450)W

= -12W

You might be interested in

adelina 88 [10]

Answer:

D

Explanation:

Michael Faraday is probably best known for his discovery of electromagnetic induction, his contributions to electrical engineering and electrochemistry or due to the fact that he was responsible for introducing the concept of field in physics to describe electromagnetic interaction.

Electromagnetic or magnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.

Electrochemistry is the branch of physical chemistry that studies the relationship between electricity, as a measurable and quantitative phenomenon, and identifiable chemical change, with either electricity considered an outcome of a particular chemical change or vice versa.

6 0

2 years ago

What is something that has a lot of thermal energy?

Artemon [7]

a heater that produces lots of warm air which would include thermal energy or a kitchen stove

8 0

2 years ago

Read 2 more answers

(a) Calculate the linear acceleration of a car, the 0.220-m radius tires of which have an angular acceleration of 11.0 rad/s2. A

Alisiya [41]

Answer:

The value is $a_t = 2.42 \ m/s^2$

Explanation:

From the question we are told that

The radius of the tires is $r = 0.22 \ m$

The angular acceleration is $\alpha = 11.0 \ rad/s^2$

Generally the linear acceleration is mathematically represented as

$a_t = r * \alpha$

=> $a_t = 0.22 * 11$

=> $a_t = 2.42 \ m/s^2$

8 0

2 years ago

What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?

VikaD [51]

(a) $1.33\cdot 10^{-11} m$

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

$e \Delta V = \frac{hc}{\lambda}$

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

$e=1.6\cdot 10^{-19}C$ is the electron's charge

$\Delta V = 93.3 kV = 93300 V$ is the potential difference

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.00\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the emitted photon

Solving the formula for $\lambda$ , we find:

$\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m$

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

$E=\frac{hc}{\lambda}$

where

h is Planck constant

c is the speed of light

$\lambda=1.33\cdot 10^{-11} m$ is the wavelength of the photon, calculated previously

Substituting,

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J$

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

$E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV$

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

$e \Delta V = \frac{hc}{\lambda}$

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0

3 years ago

An electron and a proton are each placed at rest in a uniform electric field of magnitude 560 N/C. Calculate the speed of each p

Butoxors [25]

Answer:

The speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s.

Explanation:

Given that,

Electric field, E = 560 N/C

To find,

The speed of each particle (electrons and proton) 46.0 ns after being released.

Solution,

For electron,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}$

$v=4.52\times 10^6\ m/s$

For proton,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}$

$v=2468.02\ m/s$

So, the speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.

6 0

3 years ago