A=v-v0 divided by t equals delta(the triangle) V divided by delta t
Answer:
<em>The terminal velocity at sea level is 7.99 m/s</em>
<em>The terminal velocity at an altitude of 5000 m is 10.298 m/s </em>
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Explanation:
mass of sphere m = 10 kg
radius of sphere r = 0.5 m
air density at sea level p = 1.22 kg/m^3
drag coefficient Cd = 0.8
terminal velocity = ?
Area of the sphere A = = 4 x 3.142 x = 3.142 m^2
terminal velocity is gotten from the relationship
where g = acceleration due to gravity = 9.81 m/s^2
imputing values into the equation
= <em>7.99 m/s</em>
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If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3
= <em>10.298 m/s </em>
I believe the answer is <span>216396.148556 meters per second</span>
From Newton's law v^2 = u^2 + 2as where a is the acceleration and s is the distance.
But to go any further, we need to know how fast the vehicle is accelerating
From v = u +at
We have a = u/t where the final velocity v = 0
So in one minute acceleration = (35 / 60) / 60 = 0.0097 ms/2. The first
experession in bracket is the initial velocity, u, in metres per seconds.
Hence v^2 = (0.583)^2 + 2 (0.0097)(30)
v^2 = 0.3398 + 0.5826 = 0.9224
v = âš 0.9224 = 0.960m