Answer:
The answers are below
Explanation:
a. 180 in^3 to L
1 in³ = 0.0164L
180 in³ = [tex]180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L[/tex]
b. 750 ft-lbf to kJ
1 ft-lbf = 0.00136 kJ
750 ft-lbf = [tex]750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ[/tex]
c. 75.0 hp to kW
1 hp = 0.746 kW
75 hp = [tex]75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW[/tex]
d. 2500.0 lb/h to kg/s
1 lb/h = 0.000126 kg/s
2500.0 lb/h = [tex]2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s[/tex]
e. 120 psia to kPa
1 psia = 6.89 kPa
120 psia = [tex]120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa[/tex]
f. 120 psig to kPa
1 psig = 6.89 kPa
120 psig = [tex]120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa[/tex]
g. 300 ft/min to m/s
1 ft/min = 0.005 m/s
300 ft/min = [tex]300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\[/tex]
h. 125 km/h to miles/h
1 km/h = 0.62 mph
125 km/h = [tex]125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph[/tex]
i) 6000 N to Ibf
1 N = 0.2248 lbf
6000 N = [tex]6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N[/tex]
j. 6000 N to ton
1 N = 0.000102 Ton-force
6000 N = [tex]6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N[/tex]
A brass alloy rod having a cross sectional area of 100 mm2 and a modulus of 110 GPa is subjected to a tensile load. Plastic deformation was observed to begin at a load of 39872 N. a. Determine the maximum stress that can be applied without plastic deformation. b. If the maximum length to which a specimen may be stretched without causing plastic deformation is 67.21 mm, what is the original specimen length
Answer:
a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b) length of the specimen is 66.97 mm
Explanation:
Given the data in the question;
a) Determine the maximum stress that can be applied without plastic deformation
when know that; maximum stress σ[tex]_{max}[/tex] = F / A
where F is the force in the rod ( 39872 N )
A is the cross-sectional area of the rod ( 100 mm² )
so we substitute;
σ[tex]_{max}[/tex] = 39872 N / 100 mm²
σ[tex]_{max}[/tex] = 398.72 N/mm²
Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b)
strain in the members can be calculated using the expression
ε = σ / E
where σ is the stress in the rod
E is the module of elasticity ( 110 GPa = 110000 N/mm² )
(Sl-L) / L = σ/E
where Sl-L is the change in length of the member
L is the original length of the specimen
so we substitute
(67.21 - L) / L = 398.72 / 110000
110000( 67.21 - L) = 398.72L
7393100 - 110000L = 398.72L
7393100 = 398.72L+ 110000L
7393100 = 110398.72L
L = 7393100 / 110398.72
L = 66.97 mm
Therefore; length of the specimen is 66.97 mm
Drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been
removed.
Answer:
no it has to be removed
Explanation:
It is completely inappropriate to mention that the drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been removed. Therefore, the statement given above is false.
What is the significance of drum brakes?Drum brakes can be referred to or considered as the types of brakes that are useful in application of brakes to an object, such as wheels, in motion. To understand better, it can be stated that the system of braking under drum brakes is completely in contrast to that of disc brakes.
Drum brakes have a hydraulic pressure, which means that if the condition of lining is to be checked, removal of drums becomes essential. If the drums are not removed, correction or alignment of wheels cannot be performed.
Therefore, the significance regarding drum brakes has been aforementioned, and the statement given above with respect to their removal also holds false.
Learn more about drum brakes here:
https://brainly.com/question/14937026
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Please calculate the energy stored in a flying wheel. The steel flying wheel is 10 metric tons, in adiameter of 10m, and it rotates at a speed of 1000 rpm. The steel density is 8050 kg/m^3. When energyis fully released, the flying wheel stops its rotation.
Answer:
685.38 MJ
Explanation:
Given that:
mass = 10 tons = 1.0 × 10 ⁴ kg
diameter D = 10 m
radius R = 5 m
speed N = 1000 rpm
Using the formula for K.E = [tex]\dfrac{1}{2}I \omega^2[/tex] to calculate the energy stored
where;
[tex]= \dfrac{2 \pi \times N}{60}[/tex]
[tex]= \dfrac{2 \pi \times 1000}{60}[/tex]
= 104.719 rad/s
Hence, the energy stored is;
[tex]= \dfrac{1}{2}\times (\dfrac{MR^2}{2}) \times \omega^2[/tex]
[tex]= \dfrac{1}{2}\times (\dfrac{10^4\times 5^2}{2}) \times 104.719^2[/tex]
= 685379310.1
= 685.38 MJ
Engineer drawing:
How can i draw this? Any simple way?
3.) Technician A says that a scan tool can be used to verify engine operating temperature,
Technician B says that a refractometer can be used to verify engine operating temperature.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B
64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump efficiency of 74 percent, determine the required power input to the pump. Disregard frictional losses in the pipes, and assume the geothermal water at 200m depth to be exposed to the atmosphere.
Answer:
835,175.68W
Explanation:
Calculation to determine the required power input to the pump
First step is to calculate the power needed
Using this formula
P=V*p*g*h
Where,
P represent power
V represent Volume flow rate =0.3 m³/s
p represent brine density=1050 kg/m³
g represent gravity=9.81m/s²
h represent height=200m
Let plug in the formula
P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m
P=618,030 W
Now let calculate the required power input to the pump
Using this formula
Required power input=P/μ
Where,
P represent power=618,030 W
μ represent pump efficiency=74%
Let plug in the formula
Required power input=618,030W/0.74
Required power input=835,175.68W
Therefore the required power input to the pump will be 835,175.68W
Determine the voltage drop from the top terminal to the bottom terminal, vab, in the right hand branch and, vcd, in the left hand branch of the circuit. Determine each voltage drop based on the elements in the corresponding branch.
Answer:
Hello your question is incomplete attached below is the missing part of the question
answer ;
voltage drop in the Vcd branch = 30 V
Voltage drop in the middle branch = 40v - 30v = 10 volts
voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts
Explanation:
Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch
40v and 50mA are in series hence; Ix = 50mA
also Vcd = 30V
CD is parallel to AB hence; Vcd = Vab = 30 V
Vab = ∝*Ix + 60 v
30v = ∝ ( 50mA ) + 60
therefore ∝ = -600
voltage drop in the Vcd branch = 30 V
Voltage drop in the middle branch = 40v - 30v = 10 volts
voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the following: a PWM signal source with 1 kHz base frequency and adjustable pulse-width modulation(PWM source), an analog filter with time constant of 0.01 s. Simulate the input and output of the low pass filter for a PWM duty cycle of 25, 50, and 100%
Answer:
Attached below
Explanation:
PWM signal source has 1 KHz base frequency
Analog filter : with time constant = 0.01 s
low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]
PWM duty cycle is a constant block
Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively
A Russian rocket engine (RD-110 with LOX-kerosene) consists of four thrust chambers supplied by a single turbopump. The exhaust from the turbine of the turbopump then is ducted to four vernier nozzles (which can be rotated to provide some control of the flight path). Using the information below, determine the thrust and mass flow rate of the four vernier gas nozzles. For individual thrust chambers (vacuum): F= 73.14 kN, c = 3279 m/sec Overall engine with verniers (vacuum): F= 297.93 kN, c = 3197 m/sec.
Answer:
- Vernier thrust is 5.37 kN
- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s
Explanation:
Given that;
For individual thrust chambers (vacuum);
Fc = 73.14 kN , Cc = 3279 m/sec
For Overall engine with Vernier (vacuum);
Foa = 297.93 kN = , Coa = 3197 m/sec.
- determine the Vernier thrust
Vernier thrust Fv = Foa - ( 4 × Fc )
Vernier thrust Fv = 297.93 - ( 4 × 73.14)
Vernier thrust Fv = 297.93 - 292.56
Vernier thrust Fv = 5.37 kN
Therefore, Vernier thrust is 5.37 kN
-
Vernier mass flow rate;
we know that
[tex]Co_{a}[/tex] = Fc + Fv / mc + mv
mv = Foa/Coa - Fc/Cc
we convert kilonewton to kilograms
1 kn = 102 kg
Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg
Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg
we substitute
mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)
mv = 9.5054 - 9.1006
mv = 0.4048 kg/s
Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s
why you so mean to me? leave my questions please. answer them
Answer: Why is even here then.
Explanation:
A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly Problems 49 CH001.qxd 2/24/11 12:03 PM Page 49 insulated. What is the minimum thickness of styrofoam insulation (k 0.030 W/m K) that must be applied to the top and side walls to ensure a heat load of less than 500 W, when the inner and outer surfaces are 10 and 35 C
Answer:
30 mm is the minimum thickness that must be applied.
Explanation:
Given the data in the question;
Using Fourier's equation. the heat rate is
q = kA(ΔT/Δx)
where
A is the surface area, we must consider all surfaces through which the heat can dissipate through
i.e 2×2 for one wall gives you 4m²,
there are 5 walls, so we will have 20m² for surface area.
k is thermal conductivity of the styrofoam ( 0.030 W/m K)
q is the heat loss (500 W )
ΔT is the Temperature difference ( 35 - 10) = 25°C
Δx = ?
So we substitute
500 = (0.030)(20)(25/Δx)
500 = 0.6 (25/Δx)
500 = 15 / Δx
Δx = 15 / 500
Δx = 0.03 m = 30 mm
Therefore, 30 mm is the minimum thickness that must be applied.
