Answer: more dense
Explanation:
You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?
Answer:
18 Watts
Explanation:
For this problem, we simply need to understand the relationship of power to voltage and current. This relationship is derived from Ohm's law:
Power = Voltage * Current
Given this equation, we can say the following to find the power consumption of the motor:
Power = 12volts * 1.5amps
Power = 18 Watts
Hence, the motor is consuming 18 Watts of power.
Cheers.
Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa. Determine:
a. the air-fuel ratio
b. the temperature at which the water vapor in the products will start condensing.
Answer:
44.59°c
Explanation:
Given data :
Total pressure = 105 kpa
complete combustion
A) Determine air-fuel ratio
A-F = [tex]\frac{N_{air} }{N_{fuel} } = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }[/tex]
N = number of mole
m = molar mass
A-F = [tex]\frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}[/tex] = 22.2 kg air/fuel
hence the ratio of Fuel-air = 1 : 22.2
B) Determine the temperature at which water vapor in the products start condensing
First we determine the partial pressure of water vapor before using the steam table to determine the corresponding saturation temp
partial pressure of water vapor
Pv = [tex]\frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )[/tex]
N watervapor ( number of mole of water vapor ) = 3
N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol
Pro = 105
hence Pv = ( 3/33.53 ) * 105 = 9.39kPa
from the steam pressure table the corresponding saturation temperature to 9.39kPa = 44.59° c
Temperature at which condensing will start = 44.59°c
An equation showing the products of propylene with their mole numbers is attached below
3 mA are flowing through an 18 V circuit, how much resistance (in kΩ) is in that circuit?
Answer:
6kΩ
Explanation:
If we assume that the entire circuit's current is 3 mA, then we can compute the resistance within the circuit with the application of Ohm's law:
V = IR
V/I = R
R = V / I
R = 18V / 3 mA
R = 6kΩ
Hence, the resistance of the circuit is 6kΩ.
Cheers.
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
Answer:
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
Explanation:
thats all you said
Answer:
hii my name is RAGHAV what is your name
Explanation:
this question is which chapter
What prevented this weld from becoming ropey?
A lower ampera
A higher voltage
The position of the weld
The stepping motion of the weld
Answer:
If I am not mistaken I believe it is a higher voltage.
Explanation:
Hope this helps
True or false Osha engineering controls are the last strategy of control an employer should use for job hazards
Answer:
false
Explanation:
You have a 20 Volt power source attached to a light bulb that you've measured has a resistance of 8 Ohms, what is the power output of this light bulb (in Watts)?
Answer:
50 Watts
Explanation:
For this problem, we simply apply Ohm's law:
V = IR
V / R = I
P = IV
P = ( V / R ) V
P = ( V^2 / R )
P = (20 V)^2 / 8 Ω
P = 400 V^2 / 8 Ω
P = 50 Watts
Hence, the power consumed by the lightbulb is 50 Watts.
Cheers.
If a person runs a distance of 0.7 km in 3 min, what is his average speed in kilometres/hour
Answer:
14 km/hour
Explanation:
You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the circuit?
Answer:
4mA
Explanation:
For this problem, we will simply apply Ohm's law:
V = IR
V/R = I
I = V / R
I = 12 volt / 3kΩ
I = 4mA
Hence, the current in the circuit is 4mA.
Cheers.
A flexible rectangular area measures 2.5 m X 5.0 m in plan. It supports an external load of 150 kN/m^2. Determine the vertical stress increase due to the load at a depth of 6.25 m below the corner of the footing.
Answer:
19.0476 kN/m^2
Explanation:
Given data:
Dimension of rectangular area = 2.5m x 5.0m
external load = 150 KN/m^2
load depth = 6.25 m
calculate the vertical stress increase due to load at depth 6.25
we will use the approximate method which is
[tex]V_{s} = \frac{qBL}{(B +Z)(L+Z)}[/tex] ------- (1)
q = 150 kN/m^2
B = 2.5 m
L = 5 m
Z = 6.25 m
substitute the given values into the equation (1) above
hence the Vs ( vertical stress ) = 19.0476 kN/m^2