Answer:
In fission, energy is gained by splitting apart heavy atoms, for example uranium, into smaller atoms such as iodine, caesium, strontium, xenon and barium, to name just a few. However, fusion is combining light atoms, for example two hydrogen isotopes, deuterium and tritium, to form the heavier helium.
Explanation:
I hope this helped you
(Sorry If it didn't)
Answer:
157.9 kg
Explanation:
Density: This can be defined as the ratio of the mass of a body and it's volume.
The S.I unit of density is kg/m³.
From the question,
Density = Mass/volume
D = m/v............................ Equation 1
Where D = Density of gold, m = mass of gold, v = volume of gold.
make m the subject of the equation
m = Dv.................... Equation 2
Since the gold is a cube,
v = l³................... Equation 3
Where l = length of the gold cube.
Substitute equation 3 into equation 2
m = Dl³............... Equation 4
Given: D = 19300 kg/m³, l = 0.2015 m
Substitute into equation 4
m = 19300(0.2015)³
m = 157.9 kg.
To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.
Mathematically the conservation of these two energies can be given through
![W = U_f - U_i](https://tex.z-dn.net/?f=W%20%3D%20U_f%20-%20U_i)
Where,
W = Work
Final gravitational Potential energy
Initial gravitational Potential energy
When the spacecraft of mass m is on the surface of the earth then the energy possessed by it
![U_i = \frac{-GMm}{R}](https://tex.z-dn.net/?f=U_i%20%3D%20%5Cfrac%7B-GMm%7D%7BR%7D)
Where
M = mass of earth
m = Mass of spacecraft
R = Radius of earth
Let the spacecraft is now in an orbit whose attitude is
then the energy possessed by the spacecraft is
![U_f = \frac{-GMm}{2R}](https://tex.z-dn.net/?f=U_f%20%3D%20%5Cfrac%7B-GMm%7D%7B2R%7D)
Work needed to put it in orbit is the difference between the above two
![W = U_f - U_i](https://tex.z-dn.net/?f=W%20%3D%20U_f%20-%20U_i)
![W = -GMm (\frac{1}{2R}-\frac{1}{R})](https://tex.z-dn.net/?f=W%20%3D%20-GMm%20%28%5Cfrac%7B1%7D%7B2R%7D-%5Cfrac%7B1%7D%7BR%7D%29)
Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is
![W = \frac{GMm}{2R}](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7BGMm%7D%7B2R%7D)
Answer:
1/4
Explanation:
The period of a simple pendulum is:
T = 2π √(L/g)
At half the period:
T/2
= π √(L/g)
= 2π √(L/(4g))
= 2π √((L/4)/g)
So the length would have to be shortened by a factor of 1/4.
Considering the answers.
1. Halve the voltage across it
2. Quarter the voltage across it
3. Double the voltage across it
4. Quadruple the voltage across
The appropriate answer would be 3. Double the voltage. According to ohms law the current through a conductor between two points is directly proportional to the voltage across the two points at a constant resistance. Therefore a increase in voltage causes a corresponding increase in the current.