How can a small spark start a huge explosion? using electric forces and molecules
1 answer:
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Answer:
The answer to the question is
The roller coaster will reach point B with a speed of 14.72 m/s
Explanation:
Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h
Where m = mass
g = acceleration due to gravity = 9.81 m/s²
h = starting height of the roller coaster
we have the given variables
h₁ = 36 m,
h₂ = 13 m,
h₃ = 30 m
v₁ = 1.00 m/s
Total energy at point 1 = 0.5·m·v₁² + m·g·h₁
= 0.5 m×1² + m×9.81×36
=353.66·m
Total energy at point 2 = 0.5·m·v₂² + m·g·h₂
= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m
The total energy at 1 and 2 are not equal due to the frictional force which must be considered
Total energy at point 2 = Total energy at point 1 + work done against friction
Friction work = F×d×cosθ = ( × mg)×60×cos 180 = -117.72m
0.5·m·v₂² + 127.53·m = 353.66·m -117.72m
0.5·m·v₂² = 108.41×m
v₂² = 216.82
v₂ = 14.72 m/s
The roller coaster will reach point B with a speed of 14.72 m/s
Answer:
2442.5 Nm
Explanation:
Tension, T = 8.57 x 10^2 N
length of rope, l = 8.17 m
y = 0.524 m
h = 2.99 m
According to diagram
Sin θ = (2.99 - 0.524) / 8.17
Sin θ = 0.3018
θ = 17.6°
So, torque about the base of the tree is
Torque = T x Cos θ x 2.99
Torque = 8.57 x 100 x Cos 17.6° x 2.99
Torque = 2442.5 Nm
thus, the torque is 2442.5 Nm.
