Pyramid is a type of ___________ structure.
Answer:
Massive or linteled
Explanation:
Pyramid is a type of massive or linteled structure.
These structures do no have not much internal spaces and they are huge edifices.
A pyramid is a solid body with outer triangular faces that converges on top. To construct a pyramid, large amounts of materials are usually involved. Pyramids were more prominent in times past before this present civilization.
Assume that Randy’s photocopying Service charges $.10 per photocopy. If fixed costs are
$27000 a year and variable costs are $0.04 per copy.
1. How Randy can compute his breakeven point? Show the result in graph.
2. How many photocopies are required to earn $ 500 profit?
3. Identify the safety margin at breakeven point.
Answer:
1) 739 copies (739.7260273972603).
2) 5000 copies.
Explanation:
27000/365 = 73.97260273972603 (Their daily wins). How much times $.10 is equal to 73.97260273972603?
I used a two step equation to get it.
0.10 * x = 73.97260273972603/0.10 = /0.10 x = 739.7260273972603They need to sell 739.7260273972603 copies a day to get 27000$ a year.
an equation (y = mx+b) can be
y = 739.7260273972603x
To earn a 500$ profit we need to do the same two-step equation but instead of 73.97260273972603 use 500.
0.10 * x = 500/0.10 = /0.10 x = 50005000 copies needs to be sold to get a 500$ profit.
The last question I don't understand but I hope those 2 questions helped
How do I answer all the questions on this page?
Answer:
Create a google docs copy everything and paste hope this helps! :)
Explanation:
Taking what you have learned about Ohm's law what would the current be through a circuit if the power supply was 110 v and the total resistance on the circuit was found to be 15.2 kiloOhms?
Answer:
7.24 mA
Explanation:
Ohm's law states that the current passing through two points is directly proportional to the voltage across the two points.
V ∝ I
V = IR
R is the resistance which is the constant of proportionality, V is the voltage and I is the current.
Given that V = 110 V, R = 15.2 kΩ
Therefore using ohms law, we calculate the current as:
[tex]V=IR\\\\I=\frac{V}{R}\\\\I=\frac{110}{15.2*10^3}\\\\I=7.24*10^{-3}\\\\I=7.24\ mA[/tex]
Expert Review is done by end users.
Answer:nononononono
Explanation:
Comparing cold working to hot working, flow stress is usually lower in _____ Question 19 options: Cold working Hot working
Answer:
Hot Working
Explanation:
Given that hot working is carried out at temperatures greater than the recrystallization temperature of the metal, thereby the stress needed for deformation is considerably less.
On the other hand, cold working is carried out at temperatures lesser than the recrystallization temperature of the metal, thereby stress needed for deformation is much higher
Hence, comparing cold working to hot working, flow stress is usually lower in HOT WORKING.
An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a heating value of 141,790 kJ/kg and a density of 0.0899 kg/m3, determine the efficiency of this fuel cell.
Answer:
The efficiency of this fuel cell is 80.69 percent.
Explanation:
From Physics we define the efficiency of the automotive fuel cell ([tex]\eta[/tex]), dimensionless, as:
[tex]\eta = \frac{\dot W_{out}}{\dot W_{in}}[/tex] (Eq. 1)
Where:
[tex]\dot W_{in}[/tex] - Maximum power possible from hydrogen flow, measured in kilowatts.
[tex]\dot W_{out}[/tex] - Output power of the automotive fuel cell, measured in kilowatts.
The maximum power possible from hydrogen flow is:
[tex]\dot W_{in} = \dot V\cdot \rho \cdot L_{c}[/tex] (Eq. 2)
Where:
[tex]\dot V[/tex] - Volume flow rate, measured in cubic meters per second.
[tex]\rho[/tex] - Density of hydrogen, measured in kilograms per cubic meter.
[tex]L_{c}[/tex] - Heating value of hydrogen, measured in kilojoules per kilogram.
If we know that [tex]\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}[/tex], [tex]\rho = 0.0899\,\frac{kg}{m^{3}}[/tex], [tex]L_{c} = 141790\,\frac{kJ}{kg}[/tex] and [tex]\dot W_{out} = 80\,kW[/tex], then the efficiency of this fuel cell is:
(Eq. 1)
[tex]\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)[/tex]
[tex]\dot W_{in} = 99.143\,kW[/tex]
(Eq. 2)
[tex]\eta = \frac{80\,kW}{99.143\,kW}[/tex]
[tex]\eta = 0.807[/tex]
The efficiency of this fuel cell is 80.69 percent.
A ramjet operates by taking in air at the inlet, providing fuel for combustion, and exhausting the hot air through the exit. Th e mass fl ow at the inlet and outlet of the ramjet is 60 kg/s (the mass fl ow rate of fuel is negligible). Th e inlet velocity is 225 m/s. Th e density of the gases at the exit is 0.25 kg/m3 , and the exit area is 0.5 m2 . Calculate the thr
Answer:
15300 N
Explanation:
[tex]\rho_i[/tex] = Density of air at inlet
[tex]\dfrac{m}{t}[/tex] = Mass flow rate = 60 kg/s
[tex]v_i[/tex] = Inlet velocity = 225 m/s
[tex]\rho_o[/tex] = Density of gas at outlet = [tex]0.25\ \text{kg/m}^3[/tex]
[tex]A_i[/tex] = Inlet area
[tex]A_o[/tex] = Outlet area = [tex]0.5\ \text{m}^2[/tex]
Since mass flow rate is the same in the inlet and outlet we have
[tex]\rho_iv_iA_i=\rho_ov_oA_o\\\Rightarrow v_o=\dfrac{\dfrac{m}{t}}{\rho_oA_o}\\\Rightarrow v_o=\dfrac{60}{0.25\times 0.5}\\\Rightarrow v_o=480\ \text{m/s}[/tex]
Thrust is given by
[tex]F=\dfrac{m}{t}(v_o-v_i)\\\Rightarrow F=60\times (480-225)\\\Rightarrow F=15300\ \text{N}[/tex]
The thrust generated is 15300 N.
