Answer:
Slit width is 0.109 mm
Solution:
As per the question:
Wavelength of the light,
Distance from the screen, D = 5.6 m
Bright band in the center, 2x = 8.1 cm
x = 4.05 cm
Now,
To calculate the slit width, w:
From the diffraction:
For small angle:
≈
Also
Thus
Substituting appropriate values in the above eqn:
w = 0.109 mm
Answer:
α = 3.59 [rad/s^2]]
Explanation:
First we have to convert the values of the angular speeds of revolutions per minute to radians on second.
wf = final angular velocity = 460 [rpm]
wo= initial velocity = 48 [rpm]
t = time = 12 [s]
now we can replace, in the following equation:
Answer:
μ = 0.423
Explanation:
To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body
Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis
x = v₀ t + ½ a t²
The body starts from rest so its initial speed is zero
a = 2 x / t²
a = 2 0.5 /0.5²
a = 4 m / s²
Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive
X axis
fr - Wₓ = m a (1)
Y Axis
N- = 0
N = W_{y}
We use trigonometry to find the components of the weight
sin 45 = Wₓ / W
cos 45 = W_{y} / W
Wₓ = W sin 45
W_{y} = W cos 45
The out of touch has the expression
fr = μ N
fr = μ W_{y}
We substitute in 1
μ mg cos 45 - mg sin 45 = m a
W_{y} = (a + g sin 45) / g cos 45
μ = a / g cos 45 + 1
We calculate
Acceleration goes down the plane, so it is negative
a = -4 m / s²
μ = 1- 4 / (9.8 cos 45)
μ = 0.423