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
From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.
Just as you did with #4, please sketch these on paper
as I walk you through the solutions. That'll help you see
immediately what's going on.
#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers (3 m/s) x (4 sec) = 12 meters.
Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.
The total distance he covers is (12m + 10m) = 22 meters.
#5.c).
Average speed (scalar)
= (distance covered)/(time to cover the distance)
= (22 meters)/(6 sec) = 3-2/3 m/s .
#5.d).
Displacement (vector)
= distance between the start-point and the end-point,
regardless of the route traveled,
in the direction from the start-point to the end-point.
Distance from the start-point to the end-point =
√(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters
in the direction of arctan(10/12) south of east
= 39.8° south of east.
#5.e).
Average velocity (vector) =
(displacement vector) / (time)
= 15.62 meters directed 39.8° south of east / 6 seconds
= 2.603 m/s directed 39.8° south of east.
#6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.
Direction = arctan(5.2/2.1) south of east
= 68° south of east = 158° bearing .
#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)
= 69.07 m/s .
Direction = arctan (57/39) south of west
= 55.6° south of west
Bearing = 214.4°
Compass: 0.65° past "southwest by south".
I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.
Believe it or not white clouds can be both opaque and transparent depending on the volume and mass within that cloud. The heavier the cloud, the more opaque it'll seem, but the lighter the cloud is, it'll seem transparent-
Answer:
Centripetal Acceleration = v^2/r
= (circumference/time)^2/r
= (2*pi*r/t)²)/r
= ((2³.14*50/14.3)²)/50
= 9.64 m/s²
brainlist?
Explanation:
Answer:
Yes you worked
Explanation:
You may have not seen a good outcome but you did put effort in.