Answer:
The radius of the specimen is assumed to be 9.724 mm
Explanation:
Given that:
For a cylindrical specimen of a brass alloy;
The length = 104 mm, Elongation = 5.20 mm and the tensile load = 101000 N
Let's first determine the radius of the cylindrical brass alloy from the knowledge of the cross-sectional area of a cylinder.
[tex]A_0 = \pi r ^2[/tex]
[tex]r = \sqrt{\dfrac{A_o}{\pi}}[/tex]
[tex]r = \sqrt{\dfrac{\bigg (\dfrac{F}{\sigma} \bigg )}{\pi}}[/tex]
[tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]
To estimate the tensile stress:
We need to first determine the strain relating to elongation at 5.20 mm
[tex]Strain \ \ \varepsilon= \dfrac{\Delta l}{l_o}[/tex]
[tex]Strain \ \ \varepsilon= \dfrac{5.20}{104}[/tex]
Strain ε = 0.05
Using the stress-strain plot; let assume that under the circumstances; [tex]\sigma[/tex] = 340 MPa for stress corresponding to 0.05 strain
Thus;
The cylindrical brass alloy radius [tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]
[tex]r =\sqrt{ \dfrac{101000}{(340\times 10^{6})\pi}[/tex]
r = 0.009724 m
r = 9.724 mm
A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass
of the water when the specific heat capacity of water is 4200 J/kg °C.
Answer:
2.728 kg
Explanation:
The units help you keep the calculation straight.
[tex]\dfrac{550\text{ kJ}}{(80^\circ\text{C}-32^\circ\text{C})(4.200\text{ kJ/kg\,$^\circ$C})}=\dfrac{550}{48\cdot4.2}\text{ kg}\approx\boxed{2.728\text{ kg}}[/tex]
Air enters a control volume operating at steady state at 1.2 bar, 300K, and leaves at 12 bar, 440K, witha volumetric flow rate of 1.3 m3/min. The work input to the control volume is 240 kJ per kg of air flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in kW.
Answer:
Heat transfer = 2.617 Kw
Explanation:
Given:
T1 = 300 k
T2 = 440 k
h1 = 300.19 KJ/kg
h2 = 441.61 KJ/kg
Density = 1.225 kg/m²
Find:
Mass flow rate = 1.225 x [1.3/60]
Mass flow rate = 0.02654 kg/s
mh1 + mw = mh2 + Q
0.02654(300.19 + 240) = 0.02654(441.61) + Q
Q = 2.617 Kw
Heat transfer = 2.617 Kw
1. Consider a solid cube of dimensions 1ft x 1ft x 1ft (=0.305m x 0.305m x 0.305m). Its top surface is 10
ft (=3.05 m) below the surface of the water. The density of water is pf=1000 kg/m3.
Consider two cases:
a) The cube is made of cork (pB=160.2 kg/m3)
b) The cube is made of steel (pB=7849 kg/m3)
In what direction does the body tend to move?
Answer:
a) up
b) down
Explanation:
When the cube is less dense than water, it will tend to float (move upward). When it is more dense, it will sink (move downward).
a) 160.2 kg/m^3 < 1000 kg/m^3. The cube will move up.
__
b) 7849 kg/m^3 > 1000 kg/m^3. The cube will move down.
Match the use of the magnetic field to its respective description.
oooExplanation:
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Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 2.6 kg/s. If the mixture leaves the mixing chamber at 60°C.
Required:
Determine the mass flow rate of the superheated steam required.
Answer:
0.154kg/s
Explanation:
From this question we have the following information:
P1 = 300kpa
T1 = 20⁰c
M1 = 2.6kg/s
For superheated system
P2 = 300kpa
T2 = 300⁰c
M2 = ??
T2 = 60⁰c
From saturated water table
h1 = 83.91kj/kg
h3 = 251.18kj/kg
From superheated water,
h2 = 3069.6kj/kg
The equation of energy balance
m1h1 + m2h2 = m3h3
When we input all the corresponding values:
We get
m2 = -434.902/-2818.42
m2 = 0.15430
m2 = 0.154kg/s
This is the mass flow rate of the superheated steam
Please check attachment for more detailed explanation.
thank you!
This question involves the concepts of energy balance and mass flow rate.
The mass flow rate of the superheated steam required is "0.15 kg/s".
Applying the energy balance in this situation, we get:
[tex]m_1h_1+m_2h_2=m_3h_3[/tex]
where,
m₁ = mass flow rate of liquid water at 300 KPa and 200°C = 2.6 kg/s
m₂ = mass flow rate of superheated at 300 KPa and 300°C = ?
h₁ = enthalpy of liquid water at 300 KPa and 200°C = 83.91 KJ/kg (from saturated steam table)
h₂ = enthalpy of superheated at 300 KPa and 300°C = 3069.6 KJ/kg (from superheated steam table)
h₃ = enthalpy of exiting fluid at 60°C = 251.18 KJ/kg (from saturated steam table)
m₃ = mass flow rate of exiting fluid = 2.6 kg/s + m₂
Therefore,
[tex](2.6\ kg/s)(83.91\ KJ/kg)+(m_2)(3069.6\ KJ/kg)=(2.6\ kg/s+m_2)(251.18\ KJ/kg)\\m_2(3069.6\ KJ/kg-251.18\ KJ/kg)=(2.6\ kg/s)(251.18\ KJ/kg-83.91\ KJ/kg)\\\\m_2=\frac{434.902\ KW}{2818.42\ KJ/kg}[/tex]
m₂ = 0.15 kg/s
Learn more about energy balance here:
https://brainly.com/question/9839609?referrer=searchResults
Write a program that asks the user to enter a list of numbers. The program should take the list of numbers and add only those numbers between 0 and 100 to a new list. It should then print the contents of the new list. Running the program should look something like this:
Please enter a list of numbers: 10.5 -8 105 76 83.2 206
The numbers between 0 and 100 are: 10.5 76.0 83.2
In python 3.8
nums = input("Please enter a list of numbers: ").split()
new_nums = [x for x in nums if 0 < float(x) < 100]
print("The numbers between 0 and 100 are: " + " ".join(new_nums))
When you said numbers between 0 and 100, I didn't know if that was inclusive or exclusive so I made it exclusive. I hope this helps!
A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?
Answer:
required feedback resistance ( R2 ) = 100 k Ω
Explanation:
Given data :
Voltage gain = 100
input resistance ( R1 ) = 1 k ohms
calculate feedback resistance required
voltage gain of differential amplifier
[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]
= Voltage gain = R2/R1
= 100 = R2/1
hence required feedback resistance ( R2 ) = 100 k Ω
Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
Hope this helps...........