Answer:
t = 0.24 s
Explanation:
As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:
Translation: ΣF = ma
Rotation: ΣM = Iα ; where α = angular acceleration
Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:
ΣM = I(a/R)
Now we are going to resolve and combine these equations.
For translation: Fx - Ffr = ma
We know that Fx = mgSin27°, so we substitute:
(1) mgSin27° - Ffr = ma
For rotation: (Ffr)(R) = (2/3mR²)(a/R)
The radius cancel each other:
(2) Ffr = 2/3 ma
We substitute equation (2) in equation (1):
mgSin27° - 2/3 ma = ma
mgSin27° = ma + 2/3 ma
The mass gets cancelled:
gSin27° = 5/3 a
a = (3/5)(gSin27°)
a = (3/5)(9.8 m/s²(Sin27°))
a = 2.67 m/s²
If we assume that the acceleration is a constant we can use the next equation to find the velocity:
V = √2ad; where d = 0.327m
V = √2(2.67 m/s²)(0.327m)
V = 1.32 m/s
Because V = d/t
t = d/V
t = 0.327m/1.32 m/s
t = 0.24 s
Answer
I think most see black and white.
<span>more lines = a lot of electrons returning back to ground state from same level</span>
Can you input a picture??
Answer:
4.42 x 10⁷ W/m²
Explanation:
A = energy absorbed = 500 J
η = efficiency = 0.90
E = Total energy
Total energy is given as
E = A/η
E = 500/0.90
E = 555.55 J
t = time = 4.00 s
Power of the beam is given as
P = E /t
P = 555.55/4.00
P = 138.88 Watt
d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m
Area of the circular spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Intensity of the beam is given as
I = P /A
I = 138.88 / (3.14 x 10⁻⁶)
I = 4.42 x 10⁷ W/m²
