Answer:
2 kg
Explanation:
Remember:
F = m * a re-arrange to
F/a = m substitute in the given values
10 / 5 = 2 kg
Answer:
The answer is the principal Quantum number (n)
Explanation:
The principal quantum number is one of the four quantum numbers associated with an atom.
It is denoted by a number n=1,2,3,4 etc
It tells both size (directly) and energy (indirectly) of an orbital.
When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.
It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.
Explanation:
The equation of motion of an object is given by :
Where
t is the time in seconds
We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0
So,
On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.
Answer:
The body is said to be in static equilibrium if the net force acting on a body at rest is zero.As the net force is zero,the body will not undergo motion.
Explanation:
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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