Answer:
The acceleration of
is 
Explanation:
From the question we are told that
The mass of first block is 
The angle of inclination of first block is 
The coefficient of kinetic friction of the first block is 
The mass of the second block is 
The angle of inclination of the second block is 
The coefficient of kinetic friction of the second block is 
The acceleration of
are same
The force acting on the mass
is mathematically represented as

=> 
Where T is the tension on the rope
The force acting on the mass
is mathematically represented as


At equilibrium

So

making a the subject of the formula

substituting values 
=> 
The energy absorbed by photon is 1.24 eV.
This is the perfect answer.
In the Celsius scale each degree is one part of 100 degrees. This is because in this scale the difference between boiling and freezing temperatures of water is 100 ° - 0 ° = 100 °, so one degree Celsius is one part of 100.
In the Farenheit scale, each degree is one part of 180 degrees. This is because in this scale the difference between the boiling and freezind temperatures are 212 ° - 32 ° = 180°, so one degree Farenheti is one part of 180.
That means that 1 °C is a larger amount than 1 °C, so 20°C is a larger amount than 20°F.
Conclusion: 20 degree change represents a larger change in Celsius scale.
Answer:
n1 sin θ1 = n2 sin θ2 Snell's Law (θ1 is the angle of incidence)
sin θ2 = n1 / n2 * sin θ1
sin θ2 = 2.4 / 1.33 * sin θ1
sin θ2 = 1.80 * .407 = .734
θ2 = 47.2 deg
The time of motion of the track star is determined as 0.837 s.
<h3>Time of motion of the track star</h3>
The time of motion of the track star is calculated as follows;
T = (2u sinθ)/g
where;
- T is time of motion
- g is acceleration due to gravity
- θ is angle of projection
T = (2 x 12 x sin20)/9.8
T = 0.837 s
Learn more about time of motion here: brainly.com/question/2364404
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