Answer:
Explanation:
A )
v² = u² + 2 a s
u = 0
s = 2.5
a = 9.8
v² = 2 x 9.8 x 2.5 = 49
v = 7 m /s
B )
time required to fall by first 2.5 m , t₁
s = 1/2 a t₁²
2.5 = .5 x 9.8 t₁²
t₁ = .714 s
time required to fall by 10 m
s = 1/2 a t₂²
10 = .5 x 9.8 t₂²
t₂ = 1.43 s
time required to fall by last 7.5 m
= t₂ - t₁
= 1.43 - .714
= .72 s
C )
time taken by second = .72 s
For second let initial velocity required be u
s = ut + 1/2 g t²
10 = u x .72 + .5 x 9.8 x .72²
10 = u x .72 + 2.54
u = 10.36 m /s
Answer:
dynamo generator could be the right ans
Answer:
70.5 mph
Explanation:
A passenger jet travels from Los Angeles to Bombay, India, in 22h.
The return flight takes 17 h.
The difference in flight times is caused by winds over the Pacific Ocean that
blow primarily from west to east.
If the jet's average speed in still air is 550 mi/h what is the average speed
of the wind during the round trip flight? Round to the nearest mile per hour.
Is your answer reasonable?
:
Let w = speed of the wind
:
Write a distance equation (dist is the same both ways
17(550+w) = 22(550-w)
9350 + 17w = 12100 - 22w
17w + 22w = 12100 - 9350
39w = 2750
W = 2750/39
w = 70.5 mph seems very reasonable
:
Confirming if the solution by finding the distances using these value
17(550+70.5) = 10549 mi
22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph