Answer:
Δω = -5.4 rad/s
αav = -3.6 rad/s²
Explanation:
<u>Given</u>:
Initial angular velocity = ωi = 2.70 rad/s
Final angular velocity = ωf = -2.70 rad/s (negative sign is
due to the movement in opposite direction)
Change in time period = Δt = 1.50 s
<u>Required</u>:
Change in angular velocity = Δω = ?
Average angular acceleration = αav = ?
<u>Solution</u>:
<u>Angular velocity (Δω):</u>
Δω = ωf - ωi
Δω = -2.70 - 2.70
Δω = -5.4 rad/s.
<u> Average angular acceleration (αav):</u>
αav = Δω/Δt
αav = -5.4/1.50
αav = -3.6 rad/s²
Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.
S= 343m/s
F=256Hz
WL= 343ms/256-1
WL=V/F
= 1.339844m
Answer:
d. Boyle's
Explanation:
Boyle's Law: States that the volume of a fixed mass of gas is inversely proportional proportional to its pressure, provided temperature remains constant.
Stating this mathematically. this implies that:
V∝1/P
V = k/P, Where k is the constant of proportionality
PV = k
P₁V₁ = P₂V₂
Where P₁ and P₂ are the initial and final pressure respectively, V₁ and V₂ are the the initial and final volume respectively.
Hence the right option is d. Boyle's
Answer:
20 N/m
Explanation:
From the question,
The ball-point pen obays hook's law.
From hook's law,
F = ke............................ Equation 1
Where F = Force, k = spring constant, e = compression.
Make k the subject of the equation
k = F/e........................ Equation 2
Given: F = 0.1 N, e = 0.005 m.
Substitute these values into equation 2
k = 0.1/0.005
k = 20 N/m.
Hence the spring constant of the tiny spring is 20 N/m
