Answer:

Explanation:
Given data
Terminal velocity for spread eagle position vt=130 km/h
Terminal velocity for nosedive position vt=326 km/h
The terminal speed of the diver is given by

Therefore the area is given by

Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is
Answer: 5.5km
Explanation:
Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).
Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.
Pressure decreasing with higher altitudes also means that air pressure decreases rapidly at lowerevels but more slowly at higher levels.
It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).
-2/5 = 11k - k
-2/5 = 10k
-2/5/10 = k
-2/5 * 10 = k
-2/50 = k
k = -1/25.
-1/25 - 2/5 = 11k is true.
Answer:
1.68 s
Explanation:
From newton's equation of motion,
a = (v-u)/t.................................. Equation 1
Making t the subject of the equation
t =(v-u)g............................. Equation 2
Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.
Note: Taking upward to be negative and down ward to be positive,
Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²
t = (0-8.20)/-9.8
t = -8.20/-9.8
t = 0.84 s.
But,
T = 2t
Where T = time taken for the bowling pin to return to the juggler's hand.
T = 2(0.84)
T = 1.68 s.
T = 1.68 s
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