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Determine the speed of sound in air at 300 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3

amm1812

Answer:

$c_3_0_0_K=347.19m/s$

$M=0.864$

Explanation:

The speed of sound in the air increases 0.6 m / s for every 1 ° C increase in temperature. An approximate speed can be calculated using the following empirical formula:

$c=331.5+0.6\vartheta$

Where:

$\vartheta=T-273.15K\\\\$

A more exact equation, usually referred to as adiabatic velocity of sound, is given by the following formula:

$c=\sqrt{k*R*T}$

Where:

$R= Gas\hspace{3}constant\hspace{3}of\hspace{3}air=0.287kJ/kg*K=287J/kg*K\\k=Specific\hspace{3}heat\hspace{3}ratio=1.4\\T=Temperature=300K$

Hence:

$c=\sqrt{(287)*(1.4)*(300)} =347.1887095\approx347.19m/s$

Now, the Mach number at which an aircraft is flying can be calculated by:

$M=\frac{u}{c}$

Where:

$u= Velocity\hspace{3}of\hspace{3}the\hspace{3}moving\hspace{3}aircraft\\c= Speed\hspace{3}of\hspace{3}sound\hspace{3}at\hspace{3}the\hspace{3}given \hspace{3}altitude$

Therefore:

$M=\frac{300}{347.19} =0.8640833984\approx0.864$

5 0

3 years ago

If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. Th

jonny [76]

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency

6 0

4 years ago

An automobile with a mass of 1240.00 kg has 3.99 m between

Vitek1552 [10]

Answer:

$948.15248\ N$

$5134.04751\ N$

Explanation:

$F_f$ = Force on front wheels

$F_r$ = Force on rear wheels

Distance between CG and rear wheel = 3.99-0.622 = 3.368 m

$F_r=1240\times 9.81-F_f$

As the forces are conserved we have

$(1240\times 9.81-F_f)3.368=F_f\times 0.622\\\Rightarrow 12164.4-F_f=\dfrac{0.622F_f}{3.368}\\\Rightarrow 12164.4-F_f=0.18467F_f\\\Rightarrow F_f=\dfrac{12164.4}{1.18467}\\\Rightarrow F_f=10268.09503\ N$

On rear wheels

$F_r=1240\times 9.81-10268.09503\\\Rightarrow F_r=1896.30497\ N$

The force on each rear whees is $\dfrac{1896.30497}{2}=948.15248\ N$

Force on each front wheel is $\dfrac{10268.09503}{2}=5134.04751\ N$

8 0

3 years ago

A particle is leaving the Moon in a direction that is radially outward from both the Moon and Earth.

Misha Larkins [42]

Answer:

$2780m/s$

Explanation:

Essentially, Kinetic energy of the particle must equal the combined potential energies of earth and the moon when the object is on the moon's surface, meaning the full equation is

$M_E$ =Mass of Earth= $5.97*10^2^4$

$M_m$ =Mass of Moon= $7.4*10^2^2kg$

$r_E$ =distance from earth's center to the moon's= $3.84*10^8m$

$r_m$ =radius of moon= $1.738*10^6m$

After some algebra, the equation simplifies to

$v=\sqrt{2G*(\frac{M_E}{r_E+r_m}+\frac{M_m}{r_m})}$

Plugging in the values of G, which is $6.67*10^-^1^1 \frac{m^3}{kg*s^2}$ , should yield the proper answer of 2780m/s.

4 0

3 years ago

12 seconds after starting from rest a frewly falling cantaloupe has a speed of

adoni [48]

Answer:

<em>The cantaloupe has a speed of 117.6 m/s</em>

Explanation:

<u>Free Fall Motion</u>

It occurs when an object falls under the sole influence of gravity. Any object that is being acted upon solely by the force of gravity is said to be in a state of free fall. Free-falling objects do not face air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.8 m/s^2$ .

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The cantaloupe has been dropped from rest. We are required to find the speed after t=12 seconds.

Calculate the final speed:

vf=9.8 * 12 = 117.6 m/s

The cantaloupe has a speed of 117.6 m/s

3 0

3 years ago