Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
<u>a) Calculate Natural frequency </u>
Wn = √k/m = 
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn = 
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs
Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars
Explanation :
A)
Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}
= 3780kJ
And 1 hour = 3600s
Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W
B)
At 15km/hour a 15km run takes 1 hour.
1 hour is 3600s and the runner burns 1050 joule per second.
Energy used in 1 hour = 3600 x 1050 J/s
= 3780000 J or 3.78MJ
C)
1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km
15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ
Finally,
1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ
This means that the runner needs 5320/1008 = 5.3 bars
Answer:
Follows are the solution to this question:
Explanation:
Calculating the area under the curve:
A = as

Calculating the kinematics equation:


Calculating the value of acceleration:




Answer: 3/2mg
Explanation:
Express the moment equation about point B
MB = (M K)B
-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α
α = 3g/2L cosθ
express the force equation along n and t axes.
Ft = m (aG)t
mg cosθ – Bt = m [(3g/2L cos) (L/6)]
Bt = ¾ mg cosθ
Fn = m (aG)n
Bn -mgsinθ = m[ω^2 (L/6)]
Bn =1/6 mω^2 L + mgsinθ
Calculate the angular velocity of the rod
ω = √(3g/L sinθ)
when θ = 90°, calculate the values of Bt and Bn
Bt =3/4 mg cos90°
= 0
Bn =1/6m (3g/L)(L) + mg sin (9o°)
= 3/2mg
Hence, the reactive force at A is,
FA = √(02 +(3/2mg)^2
= 3/2 mg
The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg