I'll give just one. The eruption of Mount Tambora in 1815 was one of the volcanic eruptions with widespread damage. Mount Tambora erupted, bringing thermal waves and tsunamis that killed 38,000 people (estimated). It is not just the tsunami and thermal waves, but also the ash plume that caused a cool down on the earth's atmosphere by 5°C causing "The Year Without Summer" and caused crops worldwide to fail, or degrade causing famine. That is all I know :) have a good day.
We can solve the problem by using the law of conservation of energy.
Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):

when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:

where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center

Second minima from center

The distance between the places where the intensity is zero due to the double slit effect



Put the value into the formula



Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Answer:
Explanation:
4μC will attract -9μC towards the centre and -5μC will repel it away from the centre. Both these forces are opposite to each other.
Force due to 4μC on -9μC towards the centre
= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C
Force due to -5μC on -9μC away from the centre
= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C
Ner field =407.8 N/C.
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE