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Answer:
t_total = 23.757 s
Explanation:
This is a kinematics exercise.
Let's start by calculating the distance and has to reach the limit speed of
v = 18.8 m / s
v = v₀ + a t₁
the elevator starts with zero speed
v = a t₁
t₁ = v / a
t₁ = 18.8 / 2.40
t₁ = 7.833 s
in this time he runs
y₁ = v₀ t₁ + ½ a t₁²
y₁ = ½ a t₁²
y₁ = ½ 2.40 7.833²
y₁ = 73.627 m
This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.
x_total = x₁ + x₂
x₂ = x_total - x₁
x₂ = 373 - 73,627
x₂ = 299.373 m
this distance travels at constant speed,
v = x₂ / t₂
t₂ = x₂ / v
t₂ = 299.373 / 18.8
t₂ = 15.92 s
therefore the total travel time is
t_total = t₁ + t₂
t_total = 7.833 + 15.92
t_total = 23.757 s
Answer:4.39 s
Explanation:
Given
initial velocity
acceleration
velocity acquired by sled in time
distance traveled by sled in
distance traveled in time with velocity
----1
substitute the value of in 1
we get
thus
·The acceleration of gravity is proportional to
1 / (the square of the distance from the center) .
When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .
The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)
= 490 newtons .
At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is
(9.8 m/s²) · (1/5)² = 0.39 m/s² .
The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)
= 19.6 newtons .
Just as we expected, his weight at that distance is
(19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.