3 years ago

Question 6 of 10

Physics

1 answer:

sergejj [24]3 years ago

3 0

The answer is C.89,866. I multiplied 262x343.

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The diagram shows what happens to a system undergoing an adiabatic process.

posledela

The answer is:
B. X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.

5 0

4 years ago

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Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope wit

ankoles [38]

Answer:

The acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

Explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

$a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2$

Let a₂ is acceleration of man 1,

F = m₂a₂

$a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2$

So, the acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

7 0

3 years ago

What is the lupac name H3C – CHT the H₂ - CH - CH2 CH₂ CH₂ CH3

dimulka [17.4K]

Answer:??

Explanation:

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3 years ago

iris [78.8K]

Answer:

the answer is a time your welcome

3 0

3 years ago

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Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago