According to the law of conservation of mass does the total number of atoms change during aerobic respiration

Air at 80 kPa and 400 K enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the

PilotLPTM [1.2K]

Answer:

a) 425.6 K = 152.6 degree Celsius

b) 678.6 square centimeter

Explanation:

Initial enthalpy = Final enthalpy

m(h1 + v1^2/2) = m(h2 + v2^2/2)

(h1 + v1^2/2) = (h2 + v2^2/2)

h 2 = (h1 + v1^2/2) - (v2^2/2)

h 2 = 400.98 + (230^2 – 30^2)/2 * 10^-3

= 426.98 KJ/Kg

T2 = T1’ + (T2’ -T1’)(h2-h1’)/( h2’-h1’)

T2 = 420 + (430-420)(431.43-421.26)(426.98-421.26)/(431.43 – 421.26)

T2 = 425.6 K = 152.6 degree Celsius

Area = mRT2/P2V2

Area = (600/60*60*1) *0.287*425.6/(100*30) *10^4

Area = 678.6 square centimeter

3 0

3 years ago

One major advantage of alternating current over direct current is that you can use alternating current with a device known as a

Oksanka [162]

transformer makes use of mutual induction for its operation in which change in magnetic field in one coil due to variation of current , induces voltage in the other coil. so changing magnetic field in the primary coil is very much needed for the transformer. Alternating current is a current which varies with time , hence it is suitable to produce changing magnetic field in the primary coil. on the other hand , the direct current remains constant all the time. hence can not produce a changing magnetic field. so DC current is not useful for transformers.

6 0

3 years ago

Read 2 more answers

We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,

larisa [96]

Answer:

The final charges of each sphere are: q_A = 3/8 Q , q_B = 3/8 Q , q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

q_A = Q / 2

q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

q_A = ½ (Q / 2) = ¼ Q

q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

q_A = 3/8 Q

q_B = 3/8 Q

The final charges of each sphere are:

q_A = 3/8 Q

q_B = 3/8 Q

q_C = 3/4 Q

7 0

3 years ago

g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located

Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is: P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P /RT ⋯ ⋯( I )

Here

m is the mass of the air,

v is the volume of the air,

P is the atmospheric pressure,

T is the standard temperature at the atmospheric pressure and

R is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101 kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here A is the area of the windmill

Expression to calculate the A is

A = π /4 d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94 k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94 k g / s ( (6m/s)²/2 )

P w = 160144.92 W × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0

3 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago