Answer:
Q = 40.1 degrees
Explanation:
Given:
- The weight of the timber W = 670 N
- Water surface level from pivot y = 2.1 m
- The specific density of water Y = 9810 N / m^3
- Dimension of timber = (0.15 x 0.15 x 0.0036) m
Find:
- The angle of inclination Q that the timber makes with the horizontal.
Solution:
- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:
F_b = Y * V_timber
F_b = 9810*0.15*0.15*x
F_b = 226.7*x N
- Take static equilibrium conditions for the timber, and take moments about the pivot:
(M)_p = 0
W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0
- Plug values in:
670*0.5*3.6 - x^2 * 0.5*226.7 = 0
x^2 = 1206 / 113.35
x = 3.26 m
- Now use the value of x and vertical height y to compute the angle of inclination to be:
sin(Q) = y / x
sin(Q) = 2.1 / 3.26
Q = sin^-1 (0.6441718)
Q = 40.1 degrees
Answer:
1050 kg
Explanation:
The formula for kinetic energy is:
KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.
We can now substitute the values we know into this equation.
KE = 472 500 J and v = 30 m/s:
472 500 = 1/2 × m × 30²
Next, we can rearrange the equation to make m the subject and solve for m:
m = 472 500 ÷ (1/2 × 30²)
m = 472 500 ÷ 450
m = 1050 kg
Hope this helps!
Answer:
<em>The 6000 lines per cm grating, will produces the greater dispersion .</em>
Explanation:
A diffraction grating is an optical component with a periodic (usually one that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.
The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.
For a plane diffraction grating, the angular positions of principle maxima is given by
(a + b) sin ∅n = nλ
where
a+b is the distance between two consecutive slits
n is the order of principal maxima
λ is the wavelength of the light
From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.
In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.
