A rocket is fired at a 45° angle, what is the direction of the horizontal velocity vector at the peak height?
1 answer:
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Answer:
Power_input = 85.71 [W]
Explanation:
To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.
where:
W = work [J] (units of Joules)
F = force [N] (units of Newton)
d = distance [m]
We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.
Now replacing:
Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.
where:
P = power [W] (units of watts)
W = work [J]
t = time = 40 [s]
The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.
Answer:
()=1913.31 N/m^2
Explanation:
given:
=0.85
=90 m/s
γ∞=1.23 kg/m^3
solution:
since outside pressure is atm pressure vaccum can be defined by ()
=√2(
)/γ∞[
-1]
()=1913.31 N/m^2