According to the article "Nuclear shapes" by Renee Lucas the nucleus's shape is mainly modified by vibrational and rotational features happening within the cell. According to the article if i read correctly "near closed shells spherical shapes prevail, while between closed shells the large number of valence nucleons in orbit with large particle angular momentum leads to nuclei with large deformations leading them to not only maintain its shape but also alloying it to work.
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Spring constant is 14.72 N/m</h2>
Explanation:
We have for a spring
Force = Spring constant x Elongation
F = kx
Here force is weight of mass
F = W = mg = 0.54 x 9.81 = 5.3 N
Elongation, x = 36 cm = 0.36 m
Substituting
F = kx
5.3 = k x 0.36
k = 14.72 N/m
Spring constant is 14.72 N/m
Answer:
(a) 3.44 x 10^-3 m^3/s
(b) 8.4 m/s
Explanation:
area of water line, A = 5.29 x 10^-3 m
number of holes, N = 15
Speed of water in line, V = 0.651 m/s
(a) Volume flow rate is given by
V = area of water line x speed of water in water line
V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s
(b) area of one hole, a = 4.13 x 10^-4 m
Let v be the velocity of water in each hole
According to the equation of continuity
A x V = a x v
5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v
v = 8.4 m/s
Answer:
Time period for Simple pendulum, 
Explanation:
The Simple Pendulum
Consider a small bob of mass 
is tied to extensible string of length 
that is fixed to rigid support. The bob is oscillating in the plane about verticle.
Let 
is the angle made by string with vertical during oscillation.
Vertical component of the force on bob, 
Negative sign shows that its opposing the motion of bob.
Taking as very small angle then, 
Let 
is the displacement made by bob from its mean position ,
then, 
so, 
........(1)
Since, pendulum is in hormonic motion,
as we know, 
where 
is the constant and 
.........(2)
From equation (1) and (2)
Since, 
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.
We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.
where,
- c: specific heat capacity of water
- m: mass of water
- ΔT: change in the temperature
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
Learn more: brainly.com/question/16104165
