Answer:
a) Under damped
Explanation:
Given that system is critically damped .And we have to find out the condition when gain is increased.
As we know that damping ratio given as follows
Where C is the damping coefficient and Cc is the critical damping coefficient.
So from above we can say that
From above relationship we can say when gain (K) is increases then system will become under damped system.
Answer:
reduced cost, faster marketing, and process transparency
Explanation:
The correct answer is reduced cost, faster marketing, and process transparency. PLM software helps coordinate tasks during implementation.
Answer:
<em>a) 42 mm</em>
<em>b) 144.4 MPa</em>
<em></em>
Explanation:
Load P = 200 kN = 200 x 10^3 N
Torque T = 1.5 kN-m = 1.5 x 10^3 N-m
maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa
diameter of shaft d = ?
From T = τ * *
substituting values, we have
1.5 x 10^3 = 100 x 10^6 x x
= 7.638 x 10^-5
d = = 0.042 m = <em>42 mm</em>
b) Normal stress = P/A
where A is the area
A = =
= 1.385 x 10^-3
Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = <em>144.4 MPa</em>
Answer:
Part A:
In W-h:
Energy Stored=1440 W-h
In Joules:
Part B:
In W-h:
Energy left=240 W-h
In Joules:
Energy left= 8.64*10^5 J
Explanation:
Part A:
We are given rating 120A-h and voltage 12 V
Energy Stored= Rating*Voltage (Gives us units W-h)
Energy Stored=120A-h*12V
Energy Stored=1440 W-h
Converting it into joules (watt=joules/sec)
Energy Stored=
Energy Stored=5184000 Joules
Part B:
Energy used by lights for 8h=150*8
Energy used by lights for 8h=1200W-h
Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h
Energy left=1440-1200
Energy left=240 W-h
Energy left=
Energy left=864000 Joules
Energy left= 8.64*10^5 J