Which is not part of the first law of thermodynamics?

A helium balloon containing 12m³ of gas at a pressure of 120kPa is released into the air. Assuming that the temperature is const

cluponka [151]

Answer:

36 m³

Explanation:

From gas laws, at a constant temperature, the product of volume and pressure are directly proportional hence expressed as p1v1=p2v2 where p is pressure while v represent volume. Subscripts 1 and 2 represent the initial and respectively. Making v2 the subject then

$v2=\frac {p1v1}{p2}$

Substituting 120 kpa for P1, 12m³ for V1 and 40 kpa for p2 then

$v2=\frac {120 Kpa\times 12}{40 kpa}=36 m^{3}$

Therefore, the volume will be 36 m³

6 0

3 years ago

Jerome is learning how the model of the atom has changed over time as new evidence was gathered. He has images of four models of

Yuki888 [10]

Answer:

Y, X, Z, W

Explanation:

Jerome must put the given models in the order Y, X, Z, W to display the development of atom from the earliest to the most recent one. 'Y' represents 'Thomson's plum pudding model' came in 1904 which was followed by the 'Rutherford's nuclear atomic model' of 1911 as represented by X. This was succeeded by the 'Bohr's electrostatic model' in 1913(as shown in model Z) and lastly, the model W which exemplifies the 'Quantum Mechanical Model' by Edwin Schordinger in 1926. Thus, the correct order is <u>Y, X, Z, W</u>.

6 0

3 years ago

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n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble

Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is $U_f -U_i = 16 *10^{-7} J$

Explanation:

From the question we are told that

The radius of the soap bubble is $R_o = 7.45 \ cm = \frac{7.45}{100} = 0.0745 \ m$

The potential of the soap bubble is $V_1 =307.0 V$

The new radius of the soap bubble is $R_1 = 0.19 * 7.45=1.4155\ cm = 0.014155 \ m$

The initial electric potential is mathematically represented as

$U_i = \frac{V_1^2 R_o }{2k }$

The final electric potential is mathematically represented as

$U_f = \frac{V_2^2 R_1 }{2k }$

The initial potential is mathematically represented as

$V_1 = \frac{kQ}{R_o}$

The final potential is mathematically represented as

$V_2 = \frac{kQ}{R_1}$

Now

$\frac{V_2}{V_1} = \frac{R_o}{R_1}$

substituting values

$\frac{V_2}{V_1} = \frac{7.45}{1.4155} = \frac{1}{0.19}$

=> $V_2 = \frac{V_1}{0.19}$

So

$U_f = \frac{V_1^2 R_2 }{2k * 0.19^2}$

Therefore

$U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }$

$U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]$

where k is the coulomb's constant with value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]$

$U_f -U_i = 16 *10^{-7} J$

8 0

3 years ago

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)