Answer:
minimum length of runway is needed for take off 243.16 m
Explanation:
Given the data in the question;
mass of glider = 700 kg
Resisting force = 3700 N one one glider
Total resisting force on both glider = 2 × 3700 N = 7400 N
maximum allowed tension = 12000 N
from the image below, as we consider both gliders as a system
Equation force in x-direction
2ma = T -f
a = T-f / 2m
we substitute
a = (12000 - 7400 ) / (2 × 700 )
a = 4600/1400
a = 3.29 m/s²
Now, let Vf be the final speed and Ui = 0 ( as starts from rest )
Vf² = Ui² + 2as
solve for s
Vf² = 0 + 2as
2as = Vf²
s = Vf² / 2a
given that take of speed for the gliders and the plane is 40 m/s
we substitute
s = (40)² / 2×3.29
s = 1600 / 6.58
s = 243.16 m
Therefore, minimum length of runway is needed for take off 243.16 m
Answer:
With Boyle's law we have that for a constant temperature and gas quantity the pressure of a gas multiplied by its volume is also constant: This means that under the same temperature, two gases with equal quantity of molecules and equal volume must also have the same pressure, as well as that two gases with equal quantity and pressure must have the same volume.
Explanation:
The ball has an initial speed of 10m/s. This is because it is moving with the balloon. Now the balloonist throws the ball 4m/s with respect to himself, so it means that he gives the ball a extra push of 4m/s, so the total speed is 14m/s. Since it takes 30 seconds to reach the ground, the distance travelled is 14*30=420m.
Answer:
Reverses its direction of travel precisely as it reaches the eggs
Explanation:
At the top of the motion: The spring is stretched the least, so, the potential energy of the spring is at a minimum. The mass is lifted as high as it will go, so, the potential energy due to gravity is at a maximum.
When the spring is now at the maximum, now, the maximum potential at the maximum height is
Equal to the energy stored in spring
And the energy stored in spring.
The net force on the object can be described by Hooke’s law, and so the object undergoes simple harmonic motion. Note that the initial position has the vertical displacement at its maximum value X; v is initially zero and then negative as the object moves down; and the initial acceleration is negative, back toward the equilibrium position and becomes zero at that point.
So this apply to the body given, when it get to the egg and it is released the weight will move upward 1m above the equilibrium point and it will return downward 1m below the equilibrium point, that is reverses its direction of travel precisely as it reaches the eggs.
