A skier halfway between the top and bottom of a hill. Which statement best describes the skier?

Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the

nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

Explanation:

Given Data

Initial pressure $P_{1}$ = 100 k pa

Initial temperature $T_{1}$ = 25 degree Celsius = 298 Kelvin

Final pressure $P_{2}$ = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ $V_{1}$ = $V_{2}$ ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ $\frac{P_{2} }{P_{1}}$ = $\frac{T_{2} }{T_{1}}$

⇒ Put all the values in the above formula we get the final temperature

⇒ $T_{2}$ = $\frac{300}{100}$ × 298

⇒ $T_{2}$ = 894 Kelvin

(A). Work done during the process is given by W = P × ( $V_{2} -V _{1}$ )

From equation (1), $V_{1}$ = $V_{2}$ so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m $C_{v}$ ( $T_{2}$ - $T_{1}$ )

Where m = mass of the gas = 1 kg

$C_{v}$ = specific heat at constant volume of nitrogen = 0.743 $\frac{KJ}{kg k}$

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 $\frac{KJ}{kg}$

Therefore the value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

6 0

3 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago

An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f

sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

$f=\frac{v_s}{2L}$ (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

$f=\frac{343m/s}{2(0.47228m)}=362.36Hz$

Closed tube:

$f'=\frac{v_s}{4L'}$

L': length of the closed tube = 0.702821m

$f'=\frac{343m/s}{4(0.702821m)}=122.00Hz$

Next, you use the following formula for the beat frequency:

$f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz$

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0

3 years ago

A toy car is placed at 0 on a number line. It moves 9 cm to the left, then 4 cm to the right, and then 6 cm to the len

MA_775_DIABLO [31]

Answer:19 cm

Explanation:

6 0

2 years ago

Read 2 more answers

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Pachacha [2.7K]

Answer:

d) g/2

Explanation:

We need to use one of Newton's equations of motion to find the position of the stone at any time t.

x(t) = x₀(t) + ut - ¹/₂at²

Where

x₀(t) = initial position of the stone.

x(t) - x₀(t) = distance traveled by the stone at any time.

u = initial velocity of the stone

a = acceleration of the stone

t = time taken

On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.

=> 0 = x₀(t)

=> x₀(t) = 0

On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.

=> 0 = 0 + uT - ¹/₂gT² (a = g)

=> uT = ¹/₂gT²

=> g = 2u/T

On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.

=> 0 = 0 + u(2T) - ¹/₂a(2T)²

=> 2uT = 2aT²

=> a = u/T

By comparing we see that a = g/2.

5 0

3 years ago