What is the approximate frequency of a photon having an energy 5 × 10−24 j? (h = 6.626 × 10−34 j⋅s)?
1 answer:
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Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
Answer:
r = 3519.55 m
Explanation:
We know that the acceleration of a particle in a circular motion is directed towards the center of the circumference and has magnitude:
F = rω^2
Where r is the radius of the circumference and ω is the angular velocity.
From the two acceleration vectors we find that their magnitude is
√(7^2+6^2) = √85
Therefore:
√85 m/s^2= rω^2
Now we need to calculate the angular velocity to obtain the radius. Since t2-t1 = 3s is less than one period we can be sure that the angular velocity is equals to the angle traveled between this time divided by 3 s.
The angle with respect to the x-axis for the particle at t1 and t2 is:
Therefore, the angular velocity ω is (in radians per second):
Therefore:
r = √85 / (0.0511813)^2 = 3519.55 m