Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.
Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:
Pcos 15°-N=0
Psin15°-f= m*ac
from the first we obtain N, the normal force
N=750Kg*9.8* cos (15°)= 7.1 *10^3 N
Then to calculate the frictional force (f) we can use the second equation
f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r
f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N
Answer: 2561.7 pounds
Explanation:
If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:
(1)
Where:
is the slope of the line
is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)
is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)
This means we already have one point of the graph, which coordinate is:
Rewritting (1):
(2)
As Y is a function of X:
(3)
Substituting the known values:
(4)
(5)
(6)
Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):
(7)
(8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.
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