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N76 [4]
3 years ago
12

A system absorbed 44 joules of heat from its surroundings. After doing work, the increase in the internal energy of the system i

s 32 joules. How much energy was used by the system to do useful work?
Physics
2 answers:
Minchanka [31]3 years ago
7 0

12 j is correct on edge

Lelu [443]3 years ago
4 0
We can solve the problem by using the first law of thermodynamics:
\Delta U = Q-W
where
\Delta U is the variation of internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In this problem, the heat absorbed by the system is Q=+44 J, and the increase in internal energy is \Delta U = +32 J. So we can rearrange the equation to calculate the work done:
W=Q-\Delta U=44 J-32 J=+12 J
and the work is positive, which means that it is done by the system on the surrounding.
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Answer:

<u>The transformation of energy in a torch light is as follows:</u>

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2) The electrical energy is converted into heat and light energy. (We feel the torch to be hot after some time and we can see the light energy)

Hope this helped!

<h2>~AnonymousHelper1807</h2>
7 0
3 years ago
Look at this picture of a whale shark. Which question about the whale shark is nonscientific?
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Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a
Nonamiya [84]

Answer:

1486.5\frac{Btu}{s}

Explanation:

The inlet specific volume of air is given by:

v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \  \ \  \ \ \...i

The mass flow rates is expressed as:

\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}

The energy balance for the system can the be expresses in the rate form as:

E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}

Hence, the mass flow rate of the air is 1486.5Btu/s

5 0
2 years ago
The period of a satellite circling planet Nutron is observed to be 84 s when it is in a circular orbit with a radius of 8.0 x 10
antoniya [11.8K]

Answer:

4.3 * 10^28 kg

Explanation:

Given:

Period, T = 84s

Radius of satellite orbit, r = 8*10^6

Using the relation :

M = 4π²r³ / GT²

Where G = Gravitational constant, 6.67 * 10^-11

M = 4*π^2*(8*10^6)^3 / 6.67 * 10^-11 * 84^2

M = (20218.191872 * 10^18) / 47063.52 * 10^-11

M = 0.4295937 * 10^18 - (-11)

M = 0.4295937 * 10^29

M = 4.295937 * 10^28 kg

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Mass of planet Nutron = 4.3 * 10^28 kg

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What has a definite ratio of components
Ivanshal [37]
A <span>Compound has a definte ratio of components</span>
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