Is a lump of coal chemical energy

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

A 0.18 m radius pulley is free to rotate about a horizontal axis. A mass and a mass are attached by a massless string, which is

Musya8 [376]

Answer:

T = 1.766(M-m) Nm where M and m are the 2 masses of the objects

Explanation:

Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.

Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m

T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm

8 0

3 years ago

Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.

olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

E1 = k (-q) / a²

E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

Et = E1 + E2

Et = kq / a² + kq / a²

Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

E1 = k q / (R + a)²

E2 = kq / (R-a)²

Et = kq [1 / (R + a)² + 1 / (R-a)²]

Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that R> a we can despise in the patents "a"

(R² + a²) = R² (1+ a² / R²) ≈ R²

(R + a)² = R² (1 + a / R)² ≈ R²

(R- a)² = R² (1-a / R)² ≈ R²

Substituting in the total electric field

Et = kq {2 R²) / [R²R²]}

Et =kq 2 / R²

7 0

4 years ago

Which of the following would effect the speed of a wave ? A) the matter it travels through

Artemon [7]

Answer:

Option D

The frequency

Explanation:

The speed of wave is depedant only on the wavelength and frequency of waves since it is given by s=fw where s is the speed, f is frequency and w is the wavelength. Since the options given has only one factor, that is frequency, hence option D is correct. In case we had wavelength could be among the options, both would be correct.

5 0

3 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0

3 years ago