Answer:
Spring constant, k = 5483.11 N/m
Explanation:
It is given that,
Mass of the organ, m = 2 kg
The natural period of oscillation is, T = 0.12 s
Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :



k = 5483.11 N/m
So, the spring constant for the spring in the scientist's model is 5483.11 N/m.
Answer:

Explanation:
The electrostatic attraction between the nucleus and the electron is given by:
(1)
where
k is the Coulomb's constant
Ze is the charge of the nucleus
e is the charge of the electron
r is the distance between the electron and the nucleus
This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:
(2)
where
m is the mass of the electron
v is the speed of the electron
Combining the two equations (1) and (2), we find

And solving for v, we find an expression for the speed of the electron:

Answer:
L = - 1361.591 k Kgm/s
Explanation:
Given
mA = 55.2 Kg
vA = 3.45 m/s
rA = 6.00 m
mB = 62.4 Kg
vB = 4.23 m/s
rB = 3.00 m
mC = 72.1 Kg
vC = 4.75 m/s
rC = - 5.00 m
then we apply the equation
L = (mv x r)
⇒ LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s
⇒ LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s
⇒ LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s
Finally, the total counterclockwise angular momentum of the three joggers about the origin is
L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k Kgm/s
L = - 1361.591 k Kgm/s
Answer:
4.64m/s
Explanation:
We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).
v = √2(9.8)(1.1)
v ≈ 4.64m/s
Best of Luck!