Answer:
From left = 1.2L
From back = 0.9L
Explanation:
5 x = 3(L-X) + 2(2L-X)
X = 0.7L
Distance from left = 1.2L
7y = 2(L-X)+1(2L-x)
Y = 0.4L
Distance from back = 0.9L
Answer: 
Explanation:
The Compton Shift 
in wavelength when photons are scattered is given by the following equation:
(1)
Where:
is the wavelength of the scattered photon
is the wavelength of the incident photon
is a constant whose value is given by 
, being 
the Planck constant, 
the mass of the electron and 
the speed of light in vacuum.
the angle between incident phhoton and the scatered photon.
(2)
(3) This is the shift in wavelength
Answer:
Explanation: The planet average distance = 42300km
Kepler's 3rd Law also known as the Harmonic Law states that;
for each planet orbitting the sun, its side real period squared divided by the cube of the semi-major axis of the orbit is a constant.
A planet, mass m, orbits the sun, mass M, in a circle of radius r and a period t. The net force on the planet is a centripetal force, and is caused the force of gravity between the sun and the planet.
Please find the attached file for the solution
Sounds but not solids
hoped this helped
Answer:
a) Block 1 = 72.9kgm/s
Block 2 = 0kgm/s
b) vf = 1.31m/s
c) ∆KE = 936.36Joules
Explanation:
a) Momentum = mass× velocity
For block 1:
Momentum = 2.7×27
= 72.9kgm/s
For block 2:
Momentum = 53(0) (body is initially at rest)
= 0kgm/s
b) Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses of the block
u1 and u2 are their initial velocity
v is the common velocity
Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)
2.7(27)+53(0) = (2.7+53)v
72.9 = 55.7v
V = 72.9/55.7
Vf = 1.31m/s
c) kinetic energy = 1/2mv²
Kinetic energy of block 1 = 1/2×2.7(27)²
= 984.15Joules
Kinetic energy of block 2 before collision = 0kgm/s
Total KE before collision = 984.15Joules
Kinetic energy after collision = 1/2(2.7+53)1.31²
= 1/2×55.7×1.31²
= 47.79Joules
∆KE = 984.15-47.79
∆KE = 936.36Joules
