Assume that under certain conditions the heat loss of an object is proportional to the exposed surface area. Relate the heat loss of a cubic object with side length 6 in. to one with a side length of 12 in. Now, consider two irregularly shaped objects, such as two submarines. 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. Relate the energy needed in the actual submarine to that required by the scaled model. Specify the assumptions you have made.

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

just sum

10 + 10 = 20

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%