Answer:
Its state is in uniformly accelerated motion
Explanation:
When an object is acted upon the force of gravity only, we said that the object is in free fall.
According to Newton's second law of motion:
where F is the net force on an object, m is its mass, a its acceleration, when the net force on an object (F) is non-zero, than the object accelerates (because a is non-zero), so the object is in accelerated motion.
In case of free fall, the rate of acceleration of the object is equal to 
, the acceleration due to gravity, and it is constant. So, the object is moving by uniformly accelerated motion.
The value of the angle of the incline
at which the block starts to slide is the angle at which the component of the weight parallel to the incline becomes equal to the frictional force that keeps the block on the incline:
where the term on the left is the component of the weight parallel to the incline, and the term on the right is the frictional force, which is the product between the coefficient of friction
and the normal reaction of the incline N.
The normal reaction of the incline, N, is equal to the component of the weight perpendicular to the incline:
Therefore, the initial equation becomes
From which we find
For angles above this value, the block will start sliding down, otherwise the block will stay on the incline.
Given:
u=0 m/s
a=1.1 m/s^2
S=5 m
t=time it takes to run 5 m
Use the kinematics equation
S=ut+(1/2)at^2
=>
5=0*t+(1/2)1.1(t^2)
solve for t
t=sqrt(5*2/1.1)=3.015 seconds.
Answer:
780 J
Explanation:
W=\int _{\:0}^{50}0.624xdx
If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.
so - $KE_{max} = hc/lembda} work
threshold when KE = 0
hc/lambda = work = 1240/900=1.38 eV
b) Kemax = hc/lambda - work = 1240/640 -1.38=0.558 eV
What is photocathode?
- A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
- In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
- In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.
To learn more about Photocathode from the given link:
brainly.com/question/9861585
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