Answer:
Given: mass 1200kg
initial velocity: 4m/s
finial velocity: 10 m/s
time 3 sec
then
speed; initial velocity + final velocity/2
4+10/3
: 4.66m/s2
Hello!
First one we can use that PE=mgh so we have
4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m
Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have
g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2
Hope this helps. Any questions please ask. Thank you.
The net force acting on the airplane is 25N.
Forces acting on the paper airplane when it is in the air:
- The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
- Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
- Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
- Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.
Given.
Weight of the paper airplane, F1 = 16N
The force of air resistance, F2 = 9N
Net force = F1 + F2
Net force = 25N
Thus, the net force acting on the airplane is 25N.
Learn more about the net force here:
brainly.com/question/18109210
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The formula for speed is distance÷time. The distance in this situation is 27km and the amount of time taken to cover that distance is 5 hours. Now, we substitute the values into an equation:
Speed = 27km/5h
Speed = 5.4km/h
Therefore, Desmond's speed is 5.4km/h.
