Answer:
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.
Answer:
- de Broglie wavelength of the bullet is 4.56 x 10⁻³⁴ m
- The value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.
Explanation:
Given;
mass of the bullet, m = 1.9 g = 0.0019 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ is de Broglie wavelength of the bullet
Thus, this value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.
Answer:
P.E = 1764 Joules
Explanation:
Given the following data;
Mass = 6kg
Height = 30m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
Where,
P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Substituting into the equation, we have;
P.E = 6 * 9.8 * 30
P.E = 1764 Joules