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3 years ago

A 1500-W heater is connected to a 120-V line for 2.0 hours. How much heat energy is produced?

1 answer:

3 0

<span>A joule is equal to one watt per second and so we must find out how many seconds are in two hours. A shortcut you can use is remove the zeros and multiplying the remaining numbers before adding the the zeros back to the total of your result 6 x 6 x 2 = 72 and thus 60 x 60 x 2 = 7200 72 x 15 = 1080 and thus 7200 x 1500 = 10800000 joules one mega joule equals one million joules So we can simplify things and say 10.8 MJ</span>

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7 One lb of Refrigerant 134a contained within a piston–cylinder assembly undergoes a process from a state where the temperature

julsineya [31]

Answer:

$\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb ^ {\circ} R}$

Explanation:

A) In order to solve the table it is necessary to consult tables A11-E and A10E for refrigerant R134-a

In this way we obtain that:

$S_1 = S_f = 0.0648 \frac {BTU} {Lb. ^ {\circ} R}$

In this way,

$S_2 = S_f + x_2 (S_g-S_f) = 0.0902 + 0.5 (0.2161-0.0902)$

$S_2 = 0.15315 \frac {BTU} {Lb ^ {\circ} R}$

In this way the entropy change is,

$\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb. ^ {\°} R}$

B) Whenever entropy yields a positive result, the process can be carried out adiabatically.

7 0

3 years ago

The solid right-circular cylinder of mass 500 kg is set into torque-free motion with its symmetry axis initially aligned with th

aivan3 [116]

Answer:

30.95°

Explanation:

We need to define the moment of inertia of cylinder but in terms of mass, that equation say,

$A=\frac{1}{12}m(3r^2+l^2)$

Replacing the values we have,

$A=\frac{1}{12}(500)(3(0.5)^2+(2)^2)$ $A=197.9kg.m^2$

At the same time we can calculate the mass moment of intertia of cylinder but in an axial way, that is,

$c=\frac{1}{2}mr^2$

$c=\frac{1}{2}(500)(0.5)^2$

$c=62.5kg.m^2$

Finally we need to find the required angle between the fixed line a-a (I attached an image )

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{A}{cos\gamma})^2-A^2}{c^2}}$

Replacing the values that we have,

$\Phi = 2tan^{-1}\sqrt{\frac{(\frac{197.9}{cos5\°})^2-197.9^2}{62.5^2}}$

$\Phi = 2tan^{-1}(\sqrt{0.076634})$

$\Phi = 2tan^{-1}(0.2768)$

$\Phi = 2(15.47)$

$\Phi = 30.95\°$

4 0

4 years ago

You take a lab battery and, when it isn't connected to a circuit, you measure a voltage of 4.00-V with a voltmeter. You then con

max2010maxim [7]

Answer:

$r = 28.6 ohm$

Explanation:

As we know that when battery is in open circuit then the potential difference of the cell is known as EMF

so EMF is given as

$EMF = 4.00 V$

now when the battery is connected across a resistance of 200 ohm then there is current flowing through the battery

it is given as

$i = \frac{V}{R + r}$

$17.5 \times 10^{-3} = \frac{4}{200 + r}$

$200 + r = 228.6$

$r = 28.6 ohm$

6 0

4 years ago

Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is

Naddika [18.5K]

Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

$a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2$

Therefore, the rate of acceleration of the train is 4 m/s²

5 0

3 years ago

How will you separate mixture of metal and nonmetal objects

mojhsa [17]

I think it’s by using a magnet because metals are magnetic

7 0

3 years ago