Explanation:
Given T = 10 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (10 + 273.15) K = 283.15 K
<u>T = 283.15 K </u>
The conversion of T( °C) to T(F) is shown below:
T (°F) = (T (°C) × 9/5) + 32
So,
T (°F) = (10 × 9/5) + 32 = 50 °F
<u>T = 50 °F</u>
The conversion of T( °C) to T(R) is shown below:
T (R) = (T (°C) × 9/5) + 491.67
So,
T (R) = (10 × 9/5) + 491.67 = 509.67 R
<u>T = 509.67 R</u>
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress
= 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress 
Making thickness t the subject; we have


t = 0.0056 m
t = 5.6 mm
For longitudinal stress.



t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
Answer:
≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
=
× CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, 
We can as well say:

We can now replace
in the above equation.
Therefore,
=
× CρAv²
The A which stands as the area of the jet is given by the formula:

We can now have a new equation after substituting our A into the previous equation as:
=
× Cρ 
Substituting our data from above; we have:
=
× 
= 
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= 
= 110.990 KN
≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.
Answer:The answer is Potassium!
Explanation: This is true because each label should tell you about the available amount of a certain element. The standard order is Nitrogen-Phosphorus-Potassium. They are referred to by their standard abbreviations in the periodic table. One problem with fertilizer labels are that they are only required to disclose the amounts of macronutrients (or Nitrogen-Phosphorus-Potassium.)