Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer:
2.0 m/s/s
Explanation:
The acceleration of an object is the rate of change of velocity of the object.
Mathematically, it is given by:
where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity to change from u to v
Acceleration is a vector, so it has both a magnitude and a direction.
For the runner in this problem, we have:
u = 0 is the initial velocity (he starts from rest)
v = 8.0 m/s is the final velocity
t = 4.0 s is the time taken
Substituting, we find
Answer:
The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Explanation:
Given that,
Mass of the cart, m = 250 g = 0.25 kg
Initial velocity of the cart, u = 0.31 m/s (due right)
Mass of another cart, m' = 500 g = 0.5 kg
Initial velocity of the another cart u' = -0.22 m/s (due left)
Let p is the total momentum of the system before the collision. It is given by :
So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
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