The hoop is attached.
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is
61°.</span>
A.) We use the famous equation proposed by Albert Einstein written below:
E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s
Substituting the value:
E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules
b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
The pressure needed in PSI = Pounds of force needed divided by the cylinder Area
The Cylinder rod Area is 21.19 sq inches
Thus, the pressure= 6800/21.19
= 320.91 PSI
1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³
Therefore:
8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.
Answer: 8.06 in³=1.321 x 10⁻⁴ m³