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:
6.57 m/s
Explanation:
First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement
F=kx; F=180(.3) = 54 N
Next from Newton's second law find the acceleration of the mass.
Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²
Now use the kinematic equation for velocity (or speed)
v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.
v₀=0, since the mass is at rest before we release it
a=72 m/s² (from above)
x₀=0 as the start position already compressed
x₂=0.3m (this puts the spring back to it's natural length)
v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²
v₂=
= 6.57 m/s
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Answer:
1.30
Explanation:
To calculate the critical angle we have ti use the formula:
where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.
By taking n_2 and replacing we obtain:
hope this helps!!
