The total energy of a particle is its rest energy plus the kinetic energy. Its formula is: Et= m^2/sqrt((1-v^2)/c^2))
The rest energy is equal to the product of mass and the square of light velocity: Er=mc^2.
When the kinetics energy is twice its rest energy this holds:
Et=Er
m^2/sqrt((1-v^2)/c^2))=<span>mc^2.
</span><span>sqrt((1-v^2)/c^2))=m/c^2
</span>=> v=sqrt(3/2)c
Answer:
Because there is too smooth
Answer:
= 
= 140 m / s
Explanation:
This is a projectile launch, in this case the components of the initial velocity are seen
Since there is no rice in the X axis, so the velocity is vx, it must be constant, at all times
= = 140 m / s
The vertical component of velocity changes due to having applied the gravity fiate
= 
- g t
Answer:
Explanation:
moment of inertia of flywheel = 1/2 m R²
= .5 x 1500 x .6²
= 270 kg m²
If required angular velocity be ω
rotational kinetic energy = 1/2 I ω²
= .5 x 270 x ω² = 135 ω²
kinetic energy of bus when its velocity is 20 m/s
= 1/2 x 10000 x 20²
= 2000000 J
Given 90 % of rotational kinetic energy is converted into bus's kinetic energy
135 ω² x 0.9 = 2000000 J
ω²=16461
ω = 128.3 radian /s
b )
Let the height required be h .
Total energy of bus at the top of hill = mgh + 1/2 m v²
m ( gh + .5 v²)
= 10000 ( 9.8h + .5 x 3²)
From conservation of mechanical energy theorem
10000 ( 9.8h + .5 x 3²) = 2000000
9.8h + .5 x 3² = 200
9.8h = 195.5
h = 19.95 m .
