The equation for the de Broglie wavelength is:
<span>λ = (h/mv) √[1-(v²/c²)], </span>
<span>where h is Plank's Constant, m is the rest mass, v is velocity, and c is the velocity of light in vacuum. However, if c>>v (and it is, in this case) then the expression under the radical sign approaches 1, and the equation simplifies to: </span>
<span>λ = h/mv. </span>
<span>Substituting, (remember to convert the mass to kg, since 1 J = 1 kg·m²/s²): </span>
<span>λ = (6.63x10^-34 J·s) / (0.0459 kg) (72.0 m/s) = 2.00x10^-34 m.</span>
Answer:
KE₂ = 6000 J
Explanation:
Given that
Potential energy at top U₁= 7000 J
Potential energy at bottom U₂= 1000 J
The kinetic energy at top ,KE₁= 0 J
Lets take kinetic energy at bottom level = KE₂
Now from energy conservation
U₁+ KE₁= U₂+ KE₂
Now by putting the values
U₁+ KE₁= U₂+ KE₂
7000+ 0 = 1000+ KE₂
KE₂ = 7000 - 1000 J
KE₂ = 6000 J
Therefore the kinetic energy at bottom is 6000 J.
Acceleration = (final velocity-initial velocity)/time
5 = (v-0)/20
v = 100m/s
