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4 years ago

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:

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

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

$COP_{HP} =2$

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 $40 \mathrm{m}$