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:
5.49×10⁻⁴ lbm
Explanation:
Convert volume to m³.
V = (200 cm³) (1 m / 100 cm)³ = 0.0002 m³
Find mass in kg.
m = ρV
m = (1.24507 kg/m³) (0.0002 m³)
m = 0.000249 kg
Convert mass to lbm.
m = (0.000249 kg) (2.205 lbm/kg)
m = 0.000549 lbm
m = 5.49×10⁻⁴ lbm
Answer:
Explanation:
Let's analyze the situation presented in order to know which answer is correct.
When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.
Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.
The correct answer is:
C) Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck
Answer:
Explanation:
P = Power Output = 1000 W
r = Radius = 35000000 m
= Permittivity of free space = 
c = Speed of light = 
Intensity of Electric radiation is given by
Intensity of Electric radiation is given by
The amplitude of the electric field vector is
