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Vitek1552 [10]
4 years ago
15

Make the following prefix conversions. 600,000 cm= ?? km i need help

Physics
1 answer:
Sati [7]4 years ago
4 0

Answer:

6 km

Explanation:

A kilometer is 100,000 centimeters, so 600,000 centimeters equals 6 kilometers.

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Air expands isentropically from 2.2 MPa and 77°C to 0.4 MPa. Calculate the ratio of the initial to the final speed of sound.
djyliett [7]

Answer:

The ratio of initial to final speed of sound is given as 1.28.

Explanation:

As per the thermodynamic relation of isentropic expansion

\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}

Here

  • P_1 is the pressure at point 1 which is given as 2.2 MPa
  • T_1 is the temperature at point 1 which is given as 77 °C  or 273+77=350K
  • P_2 is the pressure at point 1 which is given as 0.4 MPa
  • T_2 is the temperature at point 2 which is to be calculated
  • k is the ratio of specific heats given as 1.4

Substituting values in the equation

                                      \frac{T_2}{350}=(\frac{0.4}{2.2})^{\frac{1.4-1}{1.4}}\\\frac{T_2}{350}=(0.18)^{0.2857}\\T_2=(0.18)^{0.2857} \times 350 \\T_2=0.61266 \times 350\\T_2=214.43 K

As speed of sound c is given as

c=\sqrt{kRT}

for initial to final values it is given as

\frac{c_i}{c_f}=\frac{\sqrt{k_1R_1T_1}}{\sqrt{k_2R_2T_2}}

As values of k and R is constant so the ratio is given as

\frac{c_i}{c_f}=\sqrt{\frac{T_1}{T_2}}

Substituting values give

\frac{c_i}{c_f}=\sqrt{\frac{350}{214.43}}\\\frac{c_i}{c_f}=\sqrt{1.63}}\\\frac{c_i}{c_f}=1.277  \approx 1.28

So the ratio of initial to final speed of sound is 1.28.

5 0
3 years ago
Define velocity??????
ira [324]

Answer:

the speed of something in a given direction.

Explanation:

3 0
3 years ago
Read 2 more answers
The thermal efficiency of a power cycle operating in a reversible manner is found to be 50%. Assuming that the same 2 thermal re
inna [77]

Answer:

Explanation:

The thermal efficiency of a Power cycle \eta = \dfrac{Q_H -Q_c}{Q_H}

where;

\eta = 50\% = 0.5

Q_H = Heat \ flow \ from \ higher \ temperature

Q_c = Heat \ flow \ from \ lower \ temperature

0.5 = \dfrac{Q_H -Q_c}{Q_H}

0.5 Q_H = Q_H - Q_c --- (1)

Q_c = 0.5 Q_H         ---- (2)

The coefficient of performance is:

COP_R = \dfrac{Q_c}{Q_H -Q_c}

let replace the value of Q_c = 0.5 Q_H   in the above equation then;

COP_R = \dfrac{0.5Q_H}{Q_H -0.5 Q_H}

COP_R = \dfrac{0.5Q_H}{0.5 Q_H}

COP_R = 1

The

On the other hand,  the heat pump

COP_{HP} = \dfrac{Q_H}{Q_H -Q_c}

By replacing equation (1) into the above equation; we have:

COP_{HP} = \dfrac{Q_H}{0.5Q_{H}}

COP_{HP} = \dfrac{1}{0.5}

COP_{HP} =2

t

5 0
3 years ago
Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If
horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen 40 \mathrm{m}

Explanation:

We know, from faraday's law-

e m f=-N \frac{\Delta \Phi}{\Delta T}

and \Phi=B . A

So, as the height increases the velocity with which it will cross the ring will also increase. (v=\sqrt{2 g h})

Given

\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})

\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}

Now, from 40 \mathrm{cm}

V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}

From equation a and b we see that velocity when dropped from 40 \mathrm{m} is 10 times greater when height is 40 \mathrm{cm} so, emf will also be 10 times less as compared to when it has fallen 40 \mathrm{m}

4 0
3 years ago
Stewart (70 Kg) is attracted to Ms. Little (60 Kg) who sits 2 m away. What is the gravitational attraction between them? G=6.67×
ikadub [295]

Happy Holidays!

We can use the following equation to solve for the gravitational force:

\large\boxed{F_g = G\frac{m_1m_2}{r^2}}

Fg = force due to gravity (N)

G = Gravitational constant

m1,m2 = masses of the objects (kg)

r = distance between the objects (m)

Plug in the given values into the equation:

F_g = (6.67*10^{-11})\frac{(70)(60)}{(2)^2}} = \boxed{7.0 * 10^{-8}N}

7 0
3 years ago
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