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kati45 [8]
3 years ago
12

2 m3 of an ideal gas are compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increase

s by 10 kJ, and 150 kJ of heat is transferred to the surroundings. How much work was done by the gas during the process?
Physics
1 answer:
rosijanka [135]3 years ago
3 0

Answer:

work done is -150 kJ

Explanation:

given data

volume v1 = 2 m³

pressure p1 = 100 kPa

pressure p2 = 200 kPa

internal energy = 10 kJ

heat is transferred  = 150 kJ

solution

we know from 1st law of thermodynamic is

Q = du +W    ............1

put here value and we get

-140 = 10 + W

W = -150 kJ

as here work done is -ve so we can say work is being done on system

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What are the things being compared in the following argument?
White raven [17]

Answer:

C

Explanation:

3 0
3 years ago
In a laboratory, it is often convenient to make measurements in centimeters and grams, but SI units are needed for calculations.
zheka24 [161]

Answer:

(a) 0.92 cm= 0.092 m.

(b) 141.64 g=0.14164 kg.

(c) 15. 8 cm³=0.0000158 m³

(d) 63.6 g/cm³= 63600 kg/m³

Explanation:

The International System of Units, abbreviated S.I., also called the International System of Measurements is a system of measurements in which its units are based on fundamental physical phenomena. The units of the S.I. They are the international reference for the indications of all measuring instruments.

The International System of Units (SI) arose from the need to unify and give coherence to a great variety of unit subsystems.

The International System of Units consists of seven basic units, also called fundamental units, which define the corresponding fundamental physical quantities and which allow any physical quantity to be expressed in terms or as a combination of them. The fundamental physical quantities are complemented by two more physical quantities, called supplementary ones.

By combining the basic units, the other units are obtained, called units derived from the International System, and which allow defining any physical quantity.

(a)  The SI unit of length is the meter. Being 1 cm = 0.01 m, then <u><em>0.92 cm= 0.092 m.</em></u>

(b)  The SI unit of mass is kg. Being 1 g = 0.001 kg, then <u><em>141.64 g=0.14164 kg.</em></u>

(c) Being 1 cm³ = 0.000001 m³, then <u><em>15. 8 cm³=0.0000158 m³</em></u>

(d) Being 1 g/cm³= 1000 kg/m³, then <u><em>63.6 g/cm³= 63600 kg/m³</em></u>

3 0
3 years ago
Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m
bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0
3 years ago
104470
sergey [27]

Answer:

Grain

Explanation:

I just got it wrong :)

5 0
3 years ago
A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at
wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

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Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0
3 years ago
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