Answer:
Quantum mechanics is a key hypothesis in material science that gives a portrayal of the actual properties of nature at the size of iotas and subatomic particles. It is the establishment of all quantum physical science including quantum science, quantum field hypothesis, quantum innovation, and quantum data science.
Explanation:
It is the greatest of issues, it is the littlest of issues. At present physicists have two separate rule books clarifying how nature functions. There is general relativity, which perfectly represents gravity and everything it overwhelms: circling planets, impacting worlds, the elements of the growing universe all in all. That is enormous. At that point there is quantum mechanics, which handles the other three powers – electromagnetism and the two atomic powers. Quantum hypothesis is very proficient at portraying what happens when a uranium molecule rots, or when singular particles of light hit a sun based cell. That is little.
Answer:
v = 54 m/s
Explanation:
Given,
The maximum height of the flight of golf ball, h = 150 m
The velocity at height h, u = 0
The velocity of the golf ball right before it hits the ground, v = ?
Using the III equations of motion
<em> v² = u² + 2gh</em>
Substituting the given values in the above equation,
v² = 0 + 2 x 9.8 x 150 m
= 2940
v = 54 m/s
Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s
Answer:
puck decelerates due to the kinetic frictional force μk mg
Explanation:
given data
total distance = 12 m
coefficient of kinetic friction = 0.28
solution
we will apply equation of motion that is
v² - u² = 2 × a × s ................1
we know acceleration will be
a = 
Then we have
Force = mass × acceleration .................2
m × = -μk mg
The puck decelerates due to the kinetic frictional force μk mg
and frictional force is negative as it opposes the motion.
so we get initial velocity of the puck which is strike.
Answer:Orbital period =21.22hrs
Explanation:
given that
mass of earth M = 5.97 x 10^24 kg
radius of a satellite's orbit, R= earth's radius + height of the satellite
6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m
Speed of satellite, v= 
where G = 6.673 x 10-11 N m2/kg2
V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)
V =10,241082.2
v= 3,200.2m/s
a) Orbital period
= 
V=
T= 2 
r/ V
= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s
=76,385.1 s
60 sec= 1min
60mins = 1hr
76,385.1s =hr
76,385.1/3600=21.22hrs
