Answer:
The kinetic energy of the bullet is 5.4 × 10³ J
Explanation:
Hi there!
The equation of kinetic energy is the following:
KE = 1/2 · m · v²
Where:
KE = kinetic energy.
m = mass of the bullet.
v = speed of the bullet.
Let´s convert the mass unit to kg so that our result is in Joules:
64 g · ( 1 kg / 1000 g) = 0.064 kg
Then, the kinetic energy will be the following:
KE = 1/2 · 0.064 kg · (411 m/s)²
KE = 5.4 × 10³ J
Answers :
1. All points of a conductor are at the same potential. - True
2. Charges prefer to be uniformly distributed throughout the volume of a conductor. - False
3 The electric field inside the conducting material is always zero. -True
4.Just outside the surface of a conductor, the electric field is always zero. - False
If you're moving, then you have kinetic energy.
If you're not at the bottom yet, then you still have
some potential energy left.
They have similar physical and chemical properties because of thier valence electrons
Answer:
The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.
Explanation:
Given that,
Mass of aircraft = 10000 kg
Speed = 620 km/h = 172.22 m/s
Altitude = 10 km = 1000 m
We calculate the change in potential energy





For g = 10 m/s²,
The change in potential energy will be 1000 MJ.
We calculate the change in kinetic energy





For g = 10 m/s²,
The change in kinetic energy will be 150 MJ.
Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.