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SashulF [63]
3 years ago
15

A cylinder containing the air comprises the systemm. Cycle is completed as follows : (i) 82000 N-m of work is done by the piston

on the air during compression stroke and 45 kJ of heat are rejected to the surroundings. (ii) During expansion stroke 100000 N-m of work is done by the air on the piston. Calculate the quantity of heat added to the system؟?​
Physics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer & Explanation:

1 N-m = 1 Joule

So 82 kJ of energy put into the system during (i).

45 kJ of heat leaves the system, so 82 kJ - 45 kJ  = 37 kJ is remaining.

(ii) requires 100 kJ of energy but only 37 kJ is available, so 100 kJ - 37 kJ = 63 kJ of heat energy must be added to the system.

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3 years ago
A calorimeter uses the temperature change of water to determine the _____ of another substance.
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A calorimeter uses the temperature change of water to determine the <u>specific heat </u> of another substance.

Explanation:

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3 years ago
A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 Ω. When running at full load on
pochemuha

Answer:

The current drawn by the motor from the line is 4.68 A.

Explanation:

Given that,

Internal resistance of the dc motor, r = 3.2 ohms

Voltage, V = 120 V

Emf in the motor, \epsilon=105\ V

We need to find the current drawn by the motor from the line. A dc motor with its rotor and field coils connected in series, applying loop rule we get :

V=\epsilon+Ir

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I=\dfrac{V-\epsilon}{r}\\\\I=\dfrac{120-105}{3.2}\\\\I=4.68\ A

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3 years ago
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

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