Answer:
Jordan has more green paints
Explanation:
Given
Required
Which paint does he have more?
For better understanding, it's better to convert both measurements to decimal.
For the green paint:
For the blue paint:
By comparison:
<em>This means that Jordan has more green paints</em>
I believe it’s c because you don’t want your gas to run real low, so I think it’s best to do it when your fuel.
Answer:
Option ‘a’ is the cheapest for this house.
Explanation:
Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.
Given:
Three methods are given to heat a particular house are as follows:
Method (a)
Through Gas, this gives energy of amount $1.33/therm.
Method (b)
Through electric resistance, this gives energy of amount $0.12/KWh.
Method (c)
Through oil, this gives energy of amount $2.30/gallon.
Calculation:
Step1
Change therm to kj in method ‘a’ as follows:
$/kj.
Step2
Change kWh to kj in method ‘b’ as follows:
$/kj.
Step3
Change kWh to kj in method ‘c’ as follows:
$/kj.
Thus, the method ‘a’ has least cost as compare to method b and c.
So, option ‘a’ is the cheapest for this house.
Answer:
The critical length of surface flaw = 6.176 mm
Explanation:
Given data-
Plane strain fracture toughness Kc = 29.6 MPa-m1/2
Yield Strength = 545 MPa
Design stress. =0.3 × yield strength
= 0.3 × 545
= 163.5 MPa
Dimensionless parameter. Y = 1.3
The critical length of surface flaw is given by
= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2
Now putting values in above equation we get,
= 1/3.14( 29.6 / 1.3 × 163.5)^2
=6.176 × 10^-3 m
=6.176 mm
Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as
Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is
Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:
So we equate the COP of our heater with COP of Carnot heater
Rearrange the equation
Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C