The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
<h3>De Broglie wavelength:</h3>
The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
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Answer:
Twice.
Explanation:
The momentum of an object is given by :
p = mv
Where
m is mass and v is the velocity
If the mass of the ball were doubled, m'=2m and v'=v=3 m/s
New momentum,
p'=m'v'
p'=2m × v
p'=2mv
or
p'=2p
So, the new momentum becomes twice the initial momentum.
Answer:
0.0006091222 m
Explanation:
q = Charge = 42 pC
V = Voltage = 620 V
= Permittivity of free space = 
Electric potential is given by (at r = R)
The radius of the drop is 0.0006091222 m
Explanation:
The power P dissipated by a heater is defined as
where V is the voltage and I is the current.
a) The current running through a 130-W heater is
b) The resistance <em>R</em><em> </em>of the heater is
where 
is our familiar Ohm's Law.
Gravity is the only one helping it.
