You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y
1 answer:
Answer:
c. 2.6 h
Explanation:
The longest time spent over dinner is the time that you have available minus the minimum possible time spent in the trip.
The time of the trip is found using:
t =
Where distance is d and velocity is v. The time will be minimum at maximum velocity. Replacing with the data we have:
Ttrip = = 8.1818 h
Tdinner = 10.8h - 8.1818 h = 2.6181h
that aproximates 2.6 h.
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When is at the end of the runway the velocity of the plane is given by the equation 
where s=1800 m is the runway length. Thus
At half runway the velocity of the plane is
Therefore at midpoint of runway the percentage of takeoff velocity is
‰
At half runway the velocity of the plane is
Therefore at midpoint of runway the percentage of takeoff velocity is
‰
Answer:
The final translational seed at the bottom of the ramp is approximately 4.84 m/s
Explanation:
The given parameters are;
The radius of the sphere, R = 2.50 cm
The mass of the sphere, m = 4.75 kg
The translational speed at the top of the inclined plane, v = 3.00 m/s
The length of the inclined plane, l = 2.75 m
The angle at which the plane is tilted, θ = 22.0°
We have;
+
=
+
K = (1/2)×m×v²×(1 + I/(m·r²))
I = (2/5)·m·r²
K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²
U = m·g·h
h = l×sin(θ)
h = 2.75×sin(22.0°)
∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75ײ + 0
7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93
∴ 77.93 ≈ 7/10 × 4.75ײ
² = 77.93/(7/10 × 4.75)
≈ √(77.93/(7/10 × 4.75)) ≈ 4.84
The final translational seed at the bottom of the ramp, ≈ 4.84 m/s.