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Diano4ka-milaya

2 months ago

10

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:

Softa [3K]2 months ago

6 0

For thorough details and necessary calculations, please refer to the attachment.

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the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

serg [3582]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

$P=1.81\ kW$

We are tasked with finding the pressure here

Applying the pressure formula

$P=F\times v$

$P=A_{1}P_{1}\times v_{1}$

Here, $A_{1}v_{1}=Q_{1}$

Where, v refers to velocity

Insert the values into the equation

$1.81 =P_{1}\times0.124\times\dfrac{1}{60}$

$P_{1}=\dfrac{1.81\times60}{0.124}$

$P_{1}=875.80\ kPa$

$P_{1}=0.875\ mPa$

Therefore, the pressure at this moment is 0.875 mPa

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3 months ago

According to a newspaper account, a paratrooper survived a training jump from 1200 ft when his parachute failed to open but prov

Softa [3030]

Let's consider a few possibilities. 1. The lowest velocity of the paratrooper would be just before hitting the ground. 2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping. Hence, we will convert 100 mi/h to ft/s: 100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec. Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to 8 s * 146.67 ft/s = 1173.36 ft. This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible. Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.

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2 months ago

A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri

Maru [3345]

The charge on the plastic cube is determined as follows.

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2 months ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

3 months ago