Answer:
The electron is a subatomic particle, symbol e⁻ or β⁻ , whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure
Explanation:
functions of electrons
and electrons being the negatively charged particles of atom. Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Answer:
568.18 N
Explanation:
From the question,
The formula for gravitational potential is given as
Ep = mgh........................ Equation 1
Where Ep = Gravitational potential, m = mass of the diver,h = Height.
But,
W = mg.................... Equation 2
Where W = weight of the diver.
Substitute equation 2 into equation 1
Ep = Wh
Make W the subject of the equation
W = Ep/h................... Equation 3
Given: Ep = 25000 J, h = 44 m
Substitute into equation 3
W = 25000/44
W = 568.18 N.
Hence the weight of the diver = 568.18 N
It will decay into Silicon-30. Because alpha particles are 2 protons and 2 neutrons with an atomic mass of 4, you minus sulfur's atomic number by 2 and get silicon. And the atomic mass is 34 - 4 which equals 30.
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:

- radius of the hill:

Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car

(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,

, so we can write:

(1)
By rearranging the equation and substituting the numbers, we find N:

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:

from which we find