How does a sound wave transfer energy to your ears?
1 answer:
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Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
Answer:
k =
b =
t =
Solution:
As per the question:
Mass of the block, m = 1000 kg
Height, h = 10 m
Equilibrium position, x = 0.2 m
Now,
The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:
where
u = initial velocity = 0
g = 10
Thus
Force on the mass is given by:
F = mg =
Also, we know that the spring force is given by:
F = - kx
Thus
Now, to find the damping constant b, we know that:
F = - bv
Now,
Time required for the platform to get settled to 1 mm or 0.001 m is given by: