Hot air enters a rectangular duct (20cm wide, 25cm high, and 5m long) at 100 kPa and 60 degrees C at an average velocity of 5 m/s. While air flows the duct, it gets cool down (loses energy) so that air leave the duct at 54 degrees C. Determine the rate of heat loss from the air under steady condition

Answers

Answer 1
Answer:

Answer:

1.57 kW

Explanation:

The rate of heat loss is given by:

q = Gm * Cp * (tfin - ti)

Where

q: rate of heat loss

Gm: mass flow

Cp: specific heat at constant pressure

The Cp of air is:

Cp = 1 kJ/(kg*K)

The mass flow is the volumetric flow divided by the specific volume

Gm = Gv / v

The volumetric flow is the air speed multiplied by the cruss section of the duct.

Gv = s * h * w (I name speed s because I have already used v)

The specific volume is obtained from the gas state equation:

p * v = R * T

60 C is 333 K

The gas constant for air is 287 J/(kg*K)

Then:

v = (R * T)/p

v = (287 * 333) / 100000 = 0.955 m^3/kg

Then, the mass flow is

Gm = s * h * w / v

And rthe heat loss is of:

q = s * h * w * Cp * (tfin - ti) / v

q = 5 * 0.25 * 0.2 * 1 * (54 - 60) / 0.955 = -1.57 kW (negative because it is a loss)


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Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response with at least 3 reasons with details regarding concepts from the units learned in this course.

Answers

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from  different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Answers

The formula for the rotational kinetic energy is

KE_(rot) = (1)/(2)(number \ of\ propellers)( I)( omega)^(2)

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

I=m L^(2)=(2.67 \ kg) (0.777 \ m)^(2) =2.07459 \ kgm^(2)

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

KE_(rot) =( (1)/(2) )(5)(2.07459 \ kgm^(2)) (60\ rad/s)^(2)

KE_(rot) =18,671.31 \ J

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> 573*2\pi *(1)/(60) =60rad/s

I=(1)/(3) mL^2=(1)/(3) (2.67kg)(0.777m)^2=0.537kgm^2

K=1/2(number of rods)(I)(ω)=(1)/(2) *(5)(0.537)(60)^2=4833J

I know it's very late, but hope this helps anyone else trying to find the answer.

Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground afterSelect........seconds.

4.0

5.0

2.4

1.6


Answer and I will give you brainiliest

Answers

Answer:

2.4

Explanation:

Hope it help mark as Brainlist

We had a homework problem in which the Arrhenius equation was applied to the blinking of fireflies. Several other natural phenomena also obey that equation, including the temperature dependent chirping of crickets. A particular species, the snowy tree cricket, has been widely studied. These crickets chirp at a rate of 178 times per minute at 25.0°C, and the activation energy for the chirping process is 53.9 kJ/mol. What is the temperature if the crickets chirp at a rate of 126 times per minute?

Answers

Answer:

Temperature = 20.35°C

Explanation:

Arrhenius equation is as follows:

k = A*exp(-Ea/(R*T)), where

k = rate of chirps

Ea = Activation Energy

R = Universal Gas Constant

T = Temperature (in Kelvin)

A = Constant

Given Data

Ea = 53.9*10^3 J/mol

R = 8.3145  J/(mol.K)

T = 273.15 + 25  K

k = 178  chirps per minutes

Calculation

Using the Arrhenius equation, we can find A,

A= 4.935x10^11

Now we can apply the same equation with the data below to find T at k=126,

k = A*exp(-Ea/(R*T))

Ea = 53.9*10^3

R = 8.3145

k = 126

T = 20.35°C

A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?

Answers

Answer:

The number is  N  = 300

Explanation:

From the question we are told that

   The  net charge is  Q =  -4.8 *10^(-17 ) \  C

Generally the charge on a electron is e = - 1.60 *10^(-19 ) \ C

Generally the number of excess electrons is mathematically represented as

      N  =  (Q)/(e)

=>  N  =  (-4.8 *10^(-17))/(-1.60 *10^(-19))

=>  N  = 300

What is the frequency of an xray with a wavelength of 2x10-10m?a) 1.5x1018hz
b) 6.67x10-19hz
c) 3x108hz
d) 1.5hz

Show calculation

Answers

Answer:

1.5 x 10¹⁸hz

Explanation:

Given parameters:

Wavelength  = 2 x 10⁻¹⁰m

Unknown:

Frequency   = ?

Solution:

To find the frequency, use the expression below;

  V  = f x wavelength

V is the speed of light = 3 x 10⁸m/s

f is the frequency

Now;

 Insert the parameters

     3 x 10⁸  = 2 x 10⁻¹⁰ x frequency

   Wavelength = (3 x 10^(8) )/(2 x 10^(-10) )    = 1.5 x 10¹⁸hz