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Mariana [72]
3 years ago
8

One way to cool a gas is to let it expand. When a certain gas under a pressure of 5.00 x 10^6 Pa at 25.°C is allowed to expand t

o 3 times its original
volume, its final pressure is 1.07 x 10^6 Pa. What is its final temperature:​
Physics
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

The final temperature is T2= 5.35°C

Explanation:

Apply the Gay-lussacs's law we have

\frac{P1}{T1} = \frac{P2}{T2}

P1, initial pressure= 5.00 x 10^6 Pa

T1, initiation temperature= 25.°C

P2, final pressure= 1.07 x 10^6 Pa

T2, final temperature= ?

\frac{5.00 * 10^6}{25} =\frac{1.07 *10^6}{T2} \\

Cross multiplying and making T2 subject of formula we have

T2 =\frac{1.07 *10^6*25}{5.00 * 10^6} \\\\T2= \frac{26.75}{5} \\T2= 5.35

T2= 5.35°C

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3 years ago
What type of energy to high frequency waves have?
babunello [35]

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Gamma rays is the correct answer.

Explanation:

8 0
3 years ago
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side. (a) H
Debora [2.8K]

Answer:

Explanation:

Given

balloon is ascending at the rate of u=12\ m/s

Balloon is at a height of s=80\ m

As the package is dropped from the balloon it must possess the same initial velocity as the balloon

using

v^2-u^2=2as  

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2=12^2+2\times 9.8\times 80

v^2=1712

v=41.37\ m/s

time taken is  given by

v=u+at

substituting values

41.37=-12+9.8\times t

as we consider downward direction as positive

t=\frac{53.37}{9.8}

t=5.44\ s

therefore time taken to reach the bottom is t=5.44 s                                            

6 0
3 years ago
An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)
weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767

T₂ = 290.15 × 3.17767 = 922.00139

\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3}  = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12

\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702

Therefore;

T_4 = \frac{1383.002}{2.702} =511.859 \ k

Q_1 = c_p(T_3-T_2)

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

c_v(T_4-T_1)  = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1  \right )r_{v}^{k-1}}

Where:

β = Cut off ratio

Plugging in the values, we get;

\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1  \right )18^{1.4-1}}= 0.5191

Therefore;

\eta = \frac{\sum Q}{Q_1}

\therefore 0.5191 = \frac{150}{Q_1}

Heat supplied = \frac{150}{0.5191}  = 288.978 \ hp

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0
3 years ago
In a crash test, a 2,500 kg car hits a concrete barrier at 13 m/s2 calculate the amount of force at which the car strikes the ba
Marta_Voda [28]

Answer:

32500N

Explanation:

Data obtained from the question include:

m (mass) = 2500 kg

a (acceleration) = 13 m/s2

F (force) =?

Force is the product of mass and acceleration. It is represented mathematically as:

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F = m x a

With the above formula, the force with which the car strikes the barrier can be obtained as follow:

F = m x a

F = 2500 x 13

F = 32500N

Therefore, the car will strike the barrier with a force of 32500N

8 0
3 years ago
Read 2 more answers
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