In a free body diagram for an object projected upwards;
- the acceleration due to gravity on the object is always directed downwards.
- the velocity of the object is always in the direction of the object's motion.
An object projected upwards is subjected to influence of acceleration due to gravity.
As the object accelerates upwards, its velocity decreases until the object reaches maximum height where its velocity becomes zero and as the object descends its velocity increases, which eventually becomes maximum before the object hits the ground.
To construct a free body diagram for this motion, we consider the following;
- the acceleration due to gravity on the object is always directed downwards
- the velocity of the object is always in the direction of the object's motion.
<u>For instance:</u>
upward motion for velocity ↑ downward motion for velocity ↓
↑ ↓
↑ ↓
acceleration due to gravity ↓
↓
↓
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If you dont wear protection while shooting a rifle it will damge your hear
when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately 25 feet.
<h3>What impact, if any, would jet fuel and aviation gasoline have on a turbine engine?</h3>
Tetraethyl lead, which is present in gasoline, deposits itself on the turbine blades. Because jet fuel has a higher viscosity than aviation gasoline, it may retain impurities with greater ease.
Once the gasoline charge has been cleared, start the engine manually or with an electric starter while cutting the ignition and using the maximum throttle.
On the final approach, the aeroplane needs to be re-trimmed to account for the altered aerodynamic forces. A substantial nose-down tendency results from the airflow producing less lift on the wings and less downward force on the horizontal stabiliser due to the reduced power and slower velocity.
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Answer:
Explanation:
Let the vertical height by which it descends be h . Let it acquire velocity of v .
1/2 mv² = mgh
v² = 2gh
As it leaves the surface of sphere , reaction force of surface R = 0 , so
centripetal force = mg cosθ where θ is the angular displacement from the vertex .
mv² / r = mg cosθ
(m/r )x 2gh = mg cosθ
2h / r = cosθ
cosθ = (r-h) / r
2h / r = r-h / r
2h = r-h
3h = r
h = r / 3
The answer to this question is a) sulfur
