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:
It is equal to the overall momentum before collision, so far no external object is involved.
Explanation:
Momentum is always conserved during collision as a rule. This is equal to the product of the mass and velocity. Thank you.
Unusual precipitation patterns
Answer:
PE = 44.1 J
Explanation:
Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)
We have:
- 1 kilograms = 1000 grams.
We convert it using a rule of 3, replacing, simplifying units and solving:
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Earth's gravity is known to be 9.8 m/s², so we have:
Data:
- m = 0.3 kg
- g = 9.8 m/s²
- h = 15 m
- PE = ?
Use formula of potencial energy:
Replace and solve:
Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.
The potential energy of the volleyball is <u>44.1 Joules.</u>
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