the oscillation of rod oa about o is defined by the relation θ= (3/pi)*(sin*pi*t), where theta and t are expressed in radians an
d seconds, respectively. collar B slides along the rod so that its distance from O is r = 6(1-e^-2t) where r and t are expressed in inches and seconds. When t =1s determine (a) the velocity of the collar, (b) the acceleration of the collar (c) the acceleration of the collar relative to the rod
1 answer:
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Answer:
The answer is " and 157.5 MPa".
Explanation:
In point A:
The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point
Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.
Replacing the equation for each condition:
People can get yo = 275 MPa with both equations and k= 15.5 Mpa .
To substitute the answer,
In point b:
The equation is
equation is:
by putting the above value in the formula we get the value that is= 157.5 MPa
Answer:
gauge pressure is 133 kPa
Explanation:
given data
initial temperature T1 = 27°C = 300 K
gauge pressure = 300 kPa = 300 × 10³ Pa
atmospheric pressure = 1 atm
final temperature T2 = 77°C = 350 K
to find out
final pressure
solution
we know that gauge pressure is = absolute pressure - atmospheric pressure so
P (gauge ) = 300 × 10³ Pa - 1 × Pa
P (gauge ) = 2 × Pa
so from idea gas equation
................1
so
P2 = 2.33 × Pa
so gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 2.33 × - 1.0 ×
gauge pressure = 1.33 × Pa
so gauge pressure is 133 kPa