Answer:
The energy in kJ is 8558.16 kJ.
Explanation:
Data presented in the problem:
Water is heated from 70 (T1) to 200 °F (T2).
Volume (V) of the water is 1 ft3.
It is required for the specific heat of water(HW), which is 1 BTU/lb°F.
First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.
M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.
.After that, we can calculate the heat required (Q).
Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)
Q = 62.4 * 130 BTU = 8112 BTU.
Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ
Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.
Finally, the energy required is 8558.16 kJ.
Answer:
The speed at point B is 5.33 m/s
The normal force at point B is 694 N
Explanation:
The length of the spring when the collar is in point A is equal to:
The length in point B is:
lB=0.2+0.2=0.4 m
The equation of conservation of energy is:
(eq. 1)
Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2
in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2
Replacing in eq. 1:
Replacing values and clearing vB:
vB = 5.33 m/s
The balance forces acting in point B is:
Fc-NB-Fs=0
Replacing values and clearing NB:
NB = 694 N
Answer:
0.21
Explanation:
This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.
We need to use chord, so, I'm assuming the length of the chord to be 0.045 m
The Area is given by the formula
Area = span * chord
Area = 0.245 * 0.045
Area = 0.011 m²
This area gotten, is what we then divide the glider weight by to get our answer.
Lee = area / weight
Lee = 0.011 / 0.0523
Lee = 0.21
Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21
Answer: For #1 I'm going to go with A because that has to do with biology
For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).
For #3 I'll say marine/maritime engineer (you can just say marine)
Hope it helps!
