Against the wind a commercial airline in south america flew 784 miles in 4 hours. with a tailwind the return trip took 3.53 hours. what was the speed of the plane in still? air

First let us assign variables,

d = distance travelled

t = time it took

v = velocity of the commercial airline

In linear physics, the equation for velocity is given as:

v = d / t

Rewriting for d:

d = v t

We know that the distance to and from south America are equal therefore:

d1 (going) = d2 (return)

Let us say that velocity of air is v3. Since going to South America, the wind is against the direction of the plane and the return trip is the opposite, therefore:

(v1 - v3) t1 = (v1 + v3) t2

(v1 – v3) 4 = (v1 + v3) 3.53

4 v1 – 4 v3 = 3.53 v1 + 3.53 v3

0.47 v1 = 7.53 v3

v1 = 16.02 v3

Since we also know that:

(v1 - v3) t1 = 784

(16.02 v3 – v3) * 4 = 784

60.085 v3 = 784

v3 = 13.05 mph

Therefore the speed of the plane in still air, v1 is:

v1 = 16.02 * 13.05

v1 = 209.03 mph           (ANSWER)

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

f=ma newton

f=100x2.5=250n