How much work is done using a 500 W microwave for 5 minutes

A food department is kept at 2128C by a refrigerator in an environment at 308C. The total heat gain to the food department is es

Vesna [10]

Answer:

2.2

Explanation:

$Q_u$ = Heat rejection in the condenser = 3300 kJ/h

$Q_L$ = Heat gain to the food department = 4800 kJ/h

Power output is given by

$W=Q_u-Q_L\\\Rightarrow W=4800-3300\\\Rightarrow W=1500\ kJ/h$

COP of a refrigerator is given by

$COP=\frac{Desired\ effect}{Work}\\\Rightarrow COP=\frac{Q_L}{W}\\\Rightarrow COP=\frac{3300}{1500}\\\Rightarrow COP=2.2$

The COP of the refrigerator is 2.2

7 0

3 years ago

What is kinematics??<br>explain!!!~

Nonamiya [84]

Answer:

» Kinematics is a mathematical physics branch that deals with vectorial motion of a body

☑ Examples;

Projectile motion
Circular motion
Free fall motion

Explanation:

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4 0

2 years ago

Read 2 more answers

Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+β .

Tanzania [10]

Answer:

7.5 × 10^9 J.

Explanation:

So, we are given the following data or parameters or information for the proper solving of the question above;

=> ''actual force exerted by the tractor beam as a function of position is given by F(x) = ax + , where a = 6.1 x 10 N/mº and B = -4.1 x 10° N''

=> " Assume the supply spacecraft had an initial kinetic energy of KE = 2.7 x 10" J "

=> "that the tractor beam force is applied on the spacecraft over a distance of 7.5 x 10^11 m away.from its beginning position at = 0.0 m"

So, we can then solve it as;

KE2 – KE1 = F(x) dr.

KE2 - 2.7 x 10^11 J = - 3.5 × 10^6 × 7.5 x 10^4.

KE2 = 7.5 × 10^9 J.

3 0

3 years ago

Read 2 more answers

Given that the rms of a helium atom at a certain temperature is 1350 m s-1, determine the rms speed of an oxygen molecule at thi

monitta

Answer:

Isko moreno vs Alfredy I will gonna go to sleep soon though you were able and willing and ready to play some games ☺️☺️

5 0

2 years ago

Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68

creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is $B= 4.2 *10^ {-6}T$ , the direction is into the page

Explanation:

From the question we are told that

The current is $i = 12.0 A$

The radius of arc bc is $r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m$

The radius of arc da is $r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m$

The length of segment cd and ab is = $l = 10cm = \frac{10}{100} = 0.10 m$

The objective of the solution is to obtain the magnetic field

Generally magnetic due to the current flowing in the arc is mathematically represented as

$B = \frac{\mu_o I}{4 \pi r}$

Here I is the current

$\mu_o$ is the permeability of free space with a value of $4\pi *10^{-7}T \cdot m/A$

r is the distance

Considering Arc da

$B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta$

Where $\theta$ is the angle the arc da makes with the center from the diagram its value is $\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad$

Now substituting values into formula for magnetic field for da

$B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]$

$= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]$

$B_{da}= 12.56*10^{-6} T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

$B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta$

Substituting value

$B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]$

$B_{bc}= 8.37*10^{-6}T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

Then the net magnetic field would be

$B = B_{da} - B{bc}$

$= 12.56*10^{-6} - 8.37*10^{-6}$

$= 4.2 *10^ {-6}T$

Since $B_{da} > B_{bc}$ then the direction of the net charge would be into the page

3 0

3 years ago