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Lubov Fominskaja [6]

2 years ago

6

Thermodynamic Processes

1 answer:

goblinko [34]2 years ago

5 0

Answer:

A. Part a is the attachment

B. total work = 10.4kj

Explanation:

$workdone=nRT1ln\frac{Vb}{Va}$

T1 = constant temperature

nRT1 = PaVa = PbVb

We write equation as

$workdone =(PaVa)ln\frac{Vb}{Va}$

5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)

$w1 = workdone =(5mpa*5L)ln\frac{10L}{5L}$

W1 = 25 ln2

W1 = 25 x 0.693

= 17.327kj

The isochoric expansion has no change in volume. So,

W2 = 0

Isothermal compression

$w3=nRT3ln\frac{Vd}{Vc}$

T3 = constant temperature

nRT3 = PcVc = PdVd

$workdone=(PcVc)ln\frac{Vd}{Vc}$

Pc = 1mpa Vc = 10L Vd = 5L

$w3=(1)(10)ln\frac{5L}{10L}$

= 10x-0.693

= -6.93kj

Isochoric compression has no change in volume. Workdone w4 = 0

Total workdone = w1 + w2 + w3 + w4

= 17.33 + 0 + (-6.93) + 0

= 10.4kj

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MakcuM [25]

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3 years ago

A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

$T = 0.81 s$

Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

$v_y = 3.91 m/s$

Part d)

Take off angle is given as

$tan\theta = \frac{3.91}{3.33}$

$\theta = 49.55 degree$

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

$\Delta y = 2.50 - 1.20$

$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(1.30)$

$v_y = 5.05 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 5.05 - 9.81 t_1$

$t_1 = 0.51 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

5 0

2 years ago

Sedaia [141]

Out put work
explanation:

3 0

2 years ago

Read 2 more answers

A parallel-plate capacitor, made of two circular plates of radius R - 10 cm, is connected in series 2 with a resistor of resista

Lera25 [3.4K]

Answer:

20

Explanation:

8 0

3 years ago

A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50

asambeis [7]

Answer:

The capacitance per unit length is $5.06\times10^{-11}\ F/m$

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

$\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}$

Put the value into the formula

$\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}$

$\lambda=6.068\times10^{-9}\ C/m$

We know that,

$\lambda=\dfrac{Q}{L}$

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

$C=\dfrac{\dfrac{Q}{L}}{\Delta V}$

$C=\dfrac{6.068\times10^{-9}}{120}$

$C=5.06\times10^{-11}\ F/m$

Hence, The capacitance per unit length is $5.06\times10^{-11}\ F/m$

7 0

3 years ago