A pelican flying along a horizontal path drops
1 answer:
Answer:
The initial speed of the pelican is 8.81 m/s.
Explanation:
Given;
height of the pelican, h = 5.0 m
horizontal distance, X = 8.9 m
The time of flight is given by;
The initial horizontal speed of the pelican is given by;
X = vₓt
vₓ = X / t
vₓ = 8.9 / 1.01
vₓ = 8.81 m/s
Therefore, the initial speed of the pelican is 8.81 m/s.
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Answer:
given,
speed of the car = 20 m/s
final speed of car = 0 m/s
distance between car and the deer = 38 m
reaction time, t = 0.5 s
deceleration of the car = 10 m/s².
a) distance between deer and car
distance travel in the reaction time
d₁ = v x t
d₁ = 20 x 0.5 = 10 m
distance travel after you apply brake
using equation of motion
v² = u² + 2 a s
0 = 20² - 2 x 10 x s
s = 20 m
total distance traveled by the car
D = d₁ + d₂
D = 20 + 10 = 30 m
distance between car and the deer = 38 m - 30 m
= 8 m
b) now, maximum speed car.
distance travel in reaction time
d₁ = s x t
d₁ = 0.5 V
distance left between them
d₂ = 38 - d₁
d₂ = 38 - 0.5 V
distance travel after you apply brake
using equation of motion
v² = u² + 2 a d₂
0 = (V)² - 2 x 10 x (38 - 0.5 V)
V² + 10 V - 760 = 0
now, solving the quadratic equation
V = 23.01 , -33.01
rejecting the negative term.
hence, maximum speed of the car could be V = 23.01 m/s
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.