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Below is the solution:
9.43 m/s
<span>F = -kx </span>
<span>k = 400 N/m </span>
<span>PE= 0.5kx^2 = 0.5mv^2 </span>
<span>solve, v=9.43 m/s</span>
Answer:
Their velocity after the impact is 20.85 m/s.
Explanation:
Given that,
Mass of falcon, 
Mass of dove, 
Initial speed of the falcon, 
Initial speed of the dove, 
We need to find the final velocity after the impact. When the falcon catches the dove, it will becomes the case of inelastic collision. The conservation of momentum will be :
So, their velocity after the impact is 20.85 m/s.
Answer:
350 ft/s²
Explanation:
First, convert mph to ft/s.
58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s
Given:
v₀ = 85.1 ft/s
v = 0 ft/s
t = 0.24 s
Find: a
v = at + v₀
a = (v − v₀) / t
a = (0 ft/s − 85.1 ft/s) / 0.24 s
a = -354 ft/s²
Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².
Answer:
The speed of man before he hits the ground is <u>23.35 m/s</u>
Explanation:
We know that:
Weight of Man - Force of Friction = Unbalanced Force
but, from Newton's 2nd Law of Motion:
unbalanced force = ma
Therefore,
W - F = ma
a = (W - F)/m
a = (mg - F)/m
where,
m = 81 kg
g = 9.8 m/s²
F = 103 N
a = [(81 kg)(9.8 m/s²) - 103 N]/81 kg
a = 8.52 m/s²
using 3rd equation of motion:
Vf² - Vi² = 2ah
here,
Vi = initial velocity = 0 m/s
Vf = Final Velocity before he hits ground = ?
Vf² - 0² = 2(8.52 m/s²)(32 m)
Vf = √545.28 m²/s²
<u>Vf = 23.35 m/s</u>
Answer:
It can be seen by the eye
Explanation:
Took the test
