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Phantasy [73]
4 years ago
9

If gas in a cylinder is maintained at a constant temperature​ T, the pressure P is related to the volume V by a formula of the f

orm P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}.
Physics
1 answer:
Mkey [24]4 years ago
6 0

The given question is incomplete. The complete question is as follows.

If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form

P = \frac{nRT}{(V - nb)} - \frac{an^2}{V^2}, in which a, b, n, and R are constants. Find \frac{dP}{dV}.

Explanation:

We will use the quotient rule for each of the two terms on the right side as follows.

        P = \frac{nRT}{V - nb} - \frac{an^{2}}{V^{2}}

\frac{dP}{dV} = \frac{0(V - nb) - nRT(1)}{(V - nb)^{2}} - \frac{0(V)^{2} - an^{2}(2V)}{V^{4}}

            = \frac{-nRT}{(V - nb)^{2}} - \frac{-2an^{2}V}{V^{4}}

            = \frac{-nRT}{(V - nb)^{2}} + \frac{2an^{2}}{V^{3}}

            = \frac{2an^{2}}{V^{3}} - \frac{nRT}{(V - nb)^{2}}

   \frac{dP}{dV} = \frac{2an^{2}}{V^{3}} - \frac{nRT}{(V - nb)^{2}}

Thus, we can conclude that the value of \frac{dP}{dV} = \frac{2an^{2}}{V^{3}} - \frac{nRT}{(V - nb)^{2}}.

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Determine the kinetic energy of the ball immediately after it is hit. (You must provide an answer before moving to the next part
earnstyle [38]

The question is incomplete. Here is the complete question.

A baseball palyer hits a 5.1 oz baseball with an initial velocity of 140ft/sat an angle of 40° with the horizontal as shown. Determine

a) The kinetic energy of the ball immediately after it is hit

b) The kinetic energy of the ball when it reaches its maximum height

c) The maximum height above the ground reached by the ball.

Answer: a) KE = 131.64 J

              b) KE = 0

              c) h = 126 ft

Explanation: <u>Kinetic</u> <u>energy</u> is the energy an object posses due to its motion. It can be calculated as KE=\frac{1}{2}mv^{2}

a) Kinetic energy's unit is Joule. So, we have to transform ounce in kg and ft/s in m/s for the units to correspond:

m = 5.1(0.02835)

m = 0.1445 kg

v = 140 ft = 42.67 m/s

Then, kinetic energy is

KE=\frac{1}{2}(0.1445)(42.67)^{2}

KE = 131.64 J

Kinetic energy immediately after the ball is hit is 131.64 J.

b) At its maximum height, the ball has its highest potential energy. Because of the law of conservation of energy, when potential energy is maximum, kinetic energy is minimum and vice-versa. So, at the maximum height, kinetic energy is 0.

c) This type of motion is <u>projectile</u> <u>motion</u>. The maximum height on a projectile motion can be determined by

v_{y}^{2}=v_{0y}^{2}-2g\Delta y

When h is maximum, v_{y}=0

Velocity of the ball has an angle with the horizontal, so initial velocity at the y-axis is

v_{0y}=v_{0}sin(\theta)

Substituting and solving

v_{y}^{2}=v_{0}^{2}sin^{2}(\theta)-2gh

0=(42.67)^{2}sin^{2}(40)-2(9.8)h

19.6h=(42.67)^{2}(0.643)^{2}

h=\frac{(1820.73)(0.4132)}{19.6}

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Transforming into ft: h = 126 ft

The maximum height above the ground reached by hte ball is 126 feet.

6 0
3 years ago
A pendulum has a mass of 0.060 kg swinging at a small angle from a light string, with a period of 1.4 s. What is the length of t
jok3333 [9.3K]

Answer:

0.5 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 0.060 kg

Period (T) = 1.4 s

Lenght (L) =?

NOTE:

1. Acceleration due to gravity (g) = 10 m/s²

2. Pi (π) = 3.14

The length of the pendulum can be obtained as follow:

T = 2π√(L/g)

1.4 = 2 × 3.14 × √(L/10)

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Divide both side by 6.28

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Take the square of both side

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L = 0.5 m

Therefore, the length of the pendulum is 0.5 m

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