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2 years ago

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The PVT behavior of a certain gas is described by the equation of state: P(V − b) = RT where b is a constant. If in addition CV

is constant, show that: (a) U is a function of T only. (b) γ = const. (c) For a mechanically reversible process, P(V − b)γ = const.

1 answer:

alexdok [17]2 years ago

6 0

Answer:

shown in the attachment

Explanation:

The detailed step by step and necessary mathematical application is as shown in the attachment.

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A 5-ft-tall person walks away from the wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How

AveGali [126]

Answer:

The rate of change of the height is - 4 ft/s

Solution:

As per the question:

Height of the person, y = 5 ft

The rate at which the person walks away, $\frac{dx}{dt} = 4\ ft/s$

Distance of the spotlight from the wall, x = 40 ft

Now,

To calculate the rate of change in the height, $\frac{dy}{dt}$ of the person when, x = 10 m:

From fig 1.

$\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]Thus[tex]\frac{BC}{AB} = \frac{PQ}{QC}$

$\frac{y}{40} = \frac{5}{x}$

xy = 200 (1)

Differentiating the above eqn w.r.t time t:

$x\frac{dy}{dt} + y\frac{dx}{dt} = 0$

Thus

$\frac{dy}{dt} = - \frac{y}{x}\frac{dx}{dt}$ (2)

From eqn (1):

When x = 10 ft

10y = 200

y = 20 ft

Using eqn (2):

$\frac{dy}{dt} = - \frac{20}{10}\times 2 = - 4\ ft$

8 0

2 years ago

A string, 68 cm long and fixed at both ends, vibrates with a standing wave that has 3 antinodes. the frequency is 180 hz. what i

REY [17]

The given situation below describes a standing wave because the string is fixed at both ends. A standing wave having three anti-nodes will have a wavelength that is two-thirds the length of the string. After getting the wavelength, this can be multiplied with the frequency to get the wave speed.

For this problem:

wave length = (2/3)(length of string: 68 cm)
wave length = (10/3 cm)

wave speed = wave length x frequency
wave speed = (10/3 cm) x (180 Hz)
wave speed = 600 cm/s or 0.6 m/s

7 0

2 years ago

Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi

Julli [10]

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given as $rate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}$

$(T_{1/2})_{fission} = 3*10^{17} y$

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission $= \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}$

rate of fission =5.89*10^3 1\Year

5 0

2 years ago

A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ

3241004551 [841]

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

$PE_i + KE_i = PE_f + KE_f$

$mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2$

now here we know that

$h_1 = 2.75 m$

$v_i = 0$

$v_f = 5.23 m/s$

now plug in all values in above equation

$mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2$

divide whole equation by mass "m"

$9.8*2.75 = 9.8*h + \frac{1}{2}*27.35$

$9.8*h = 13.27$

$h = 1.35 m$

so height of the ball from ground will be 1.35 m

4 0

1 year ago

A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0

2 years ago