Answer:
factor of 100 more heat loss in actual
Step-by-step explanation:
Given:
- Length of the cube = 6 in
- Length of another cube = 12 in
- Actual submarine size = 70 ft
- Scale model = 7-ft
Find:
Relate the heat loss of a 70-ft submarine to that of a 7-ft scale model. Suppose you are interested in the amount of energy needed to maintain a con- stant internal temperature in the submarine.
Solution:
- The surface area of the small cube is 6 · 36.
- Scaling the sides by k means the new surface area is 6 · (6k · 6k) = k
^2*
(6 · 36).
- Therefore, double the lengths increases the surface area 4 times (and so there will also be 4 times as much heat loss).
- Now consider two irregularly shaped objects, such as submarines. If we
can measure heat loss from a 7 foot scale model, heat is lost from a 70 foot sub is:
- An increase by a factor of 10 will mean that heat loss (which is proportional to surface area) increases by a factor of 10^2 = 100
Answer:
Step-by-step explanation:
-4s - 7 = 29 + 5s
Subtract 5s from both sides [-4s - 5s - 7 = 29 + 5s - 5s]
-9s - 7 = 29
Add 7 to both sides [-9s -7 +7 = 29 + 7]
-9s = 36
Divide both sides by (-9) [-9s/-9 = 36/-9]
s = -4
Let h represent number of hits.
Let m represent number of misses.
1) 9h - 7m = 33
2) h - m = 2
2a) h = 2 + m
Sub 2a) into 1)
9*(2+m) - 7m = 33
18 + 9m - 7m = 33
2m = 15
m = 7.5
Sub m = 7.5 into 2a)
h = 2+ 7.5
h = 9.5
Though this does work you can’t exactly shoot a half miss or hit so h=13 and m=12 works as well...
9(13)-7(12)=33
117- 84=33
£70
if 100%= ?
80%=56
100/80=1.25
1.25 x 56
=70