Some lactic acid in the muscles dissipates during .
1 answer:
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Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
<u>α = - 1.047 rad/s²</u>
B.
We can use the second equation of motion to find the angular distance:
<u>θ = 14.1 rad</u>
C.
θ = (14.1 rad)(1 rev/2π rad)
<u>θ = 2.24 rev</u>
The rate at which the height is changing is ( 5 / x ) m / hr
We know that,
Area of an equilateral triangle A =
/ 4
h = x / 2
Where,
x = Side
h = Height
Given that,
dA / dt = 5 / hr
h = x / 2
Differentiate both sides with respect to t
dh / dt = ( / 2 ) ( dx / dt )
dx / dt = ( 2 / ) ( dh / dt )
A =
/ 4
Differentiate both sides with respect to t
dA / dt = ( / 4 ) ( 2x ) ( dx / dt )
5 = ( / 4 ) ( 2x ) ( 2 /
) ( dh / dt )
dh / dt = ( 5 / x ) m / hr
Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt
Therefore, the rate at which the height is changing is ( 5 / x ) m / hr
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