Time in the universe vs the population
Answer:
g' = 10.12m/s^2
Explanation:
In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.
You use the following formula:
(1)
l: length of the pendulum = ?
g: acceleration due to gravity at sea level = 9.79m/s^2
T: period of the pendulum at sea level = 1.2s
You solve for l in the equation (1):
![l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m](https://tex.z-dn.net/?f=l%3D%5Cfrac%7BgT%5E2%7D%7B4%5Cpi%5E2%7D%5C%5C%5C%5Cl%3D%5Cfrac%7B%289.79m%2Fs%5E2%29%281.2s%29%5E2%7D%7B4%5Cpi%5E2%7D%3D0.35m)
Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:
![T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}](https://tex.z-dn.net/?f=T%27%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%27%7D%7D%5C%5C%5C%5Cg%27%3D%5Cfrac%7B4%5Cpi%5E2l%7D%7BT%27%5E2%7D)
g': acceleration due to gravity at the top of the mountain
T': new period of the pendulum
![g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}](https://tex.z-dn.net/?f=g%27%3D%5Cfrac%7B4%5Cpi%5E2%280.35m%29%7D%7B%281.18s%29%5E2%7D%3D10.12%5Cfrac%7Bm%7D%7Bs%5E2%7D)
The acceleration due to gravity at the top of the mountain is 10.12m/s^2
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
![U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J](https://tex.z-dn.net/?f=U_i%20%3D%20Fr%20%3D%20%5Cfrac%7Bkq%5E2%7D%7Br%5E2%7D%20%5Ctimes%20r%20%3D%20%5Cfrac%7Bkq%5E2%7D%7Br%7D%20%5C%5C%5C%5CU_i%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%281.602%5Ctimes%2010%5E%7B-19%7D%29%5E2%7D%7B0.09%7D%20%5C%5C%5C%5CU_i%20%3D%202.566%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J)
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
![U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7Bkq%5E2%7D%7Br_2%7D%20%5C%5C%5C%5CU_f%20%3D%20%20%5B%5Cfrac%7B%289%5Ctimes%2010%5E9%29%5Ctimes%20%281.602%20%5Ctimes%2010%5E%7B-19%7D%29%5E2%7D%7B0.03%7D%20%5D%5C%5C%5C%5CU_f%20%3D%207.669%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J%20%5C%5C%5C%5C%5CDelta%20U%20%3D%20U_f%20-U_i%5C%5C%5C%5C%5CDelta%20U%20%3D%20%287.699%5Ctimes%2010%5E%7B-27%7D%20%5C%20J%20%29%20-%20%282.566%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J%29%5C%5C%5C%5C%5CDelta%20U%20%3D%205.133%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20J)
Apply the principle of conservation of energy;
ΔK.E = ΔU
![K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s](https://tex.z-dn.net/?f=K.E_f%20-K.E_i%20%3D%20%5CDelta%20U%5C%5C%5C%5Cinitial%20%5C%20velocity%20%5C%20of%20%20%5C%20the%20%5C%20electron%20%3D%200%5C%5C%5C%5CK.E_f%20-%200%20%3D%20%5CDelta%20U%5C%5C%5C%5CK.E_f%20%3D%20%5CDelta%20U%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%3D%20%5CDelta%20U%5C%5C%5C%5Cwhere%3B%5C%5C%5C%5Cm%20%5C%20is%20%5C%20the%20%5C%20mass%20%5C%20of%5C%20the%20%5C%20electron%20%3D%209.1%201%20%5Ctimes%2010%5E%7B-31%7D%20%5C%20kg%5C%5C%5C%5Cv%5E2%20%3D%20%5Cfrac%7B%202%20%5CDelta%20U%7D%7Bm%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B%202%20%5CDelta%20U%7D%7Bm%7D%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B%202%20%285.133%5Ctimes%2010%5E%7B-27%7D%29%7D%7B9.11%5Ctimes%2010%5E%7B-31%7D%7D%7D%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B11268.935%7D%20%5C%5C%5C%5Cv%20%3D%20106.2%20%5C%20m%2Fs)
Therefore, the speed of the electron at the given position is 106.2 m/s
Answer:
The helicopter was deformed and destroyed in the inelastic collision.
Explanation:
- When two object collide there exist two way of colliding: elastic collision and inelastic collision.
- Two terms are considered during the collision: kinetic energy and momentum.
- If both of these terms are conserved in any collision then there is no significant loss of property, this is called as elastic collision.
- If only momentum is conserved but kinetic energy is converted into other forms then it is inelastic collision. In inelastic collision, the energy is lost in the form of vibration, sound etc. causing the damage to colliding object.
- Hence the deformation of helicopter was due to inelastic collision.
The viscosity of magma is directly related to the silica content. That being said, the viscosity of magma is higher when the silica content is also high and it is a low viscosity magma when the silica content is also low. The silica content of low viscosity magma is approximately 45 - 55 wt%.