Answer:
<em>The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>
Explanation:
The question is incomplete. So the picture, which is missing in question, is attached for your review.
As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).
<em>So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>
The current will lag the voltage in AC circuit that contains both resistance and inductance.
Answer: C
Explanation
There is no inductance only circuits in reality.
The circuits containing inductance has also a lower amount of resistance.
The current flows in both resistance and inductance.
There is a drop in the total voltage in resistance and inductance giving rise to the voltage applied in the coil when connected in a series.
An example being inductance coil an AC circuit connected to both resistance and inductance in series.
From the vector diagram, this conclusion can be drawn.
Answer:
See the detailed answer in attached file.
Explanation :
Answer:
(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.
Equation 11-1: F*L^(1/3) = Constant
Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483
Explanation:
(a)The Catalog rating(C)
Bearing life:
Catalog rating: 
From given equation bearing life equation,

we Dividing eqn (2) with (1)

The Catalog rating increased by factor of 1.26
(b) Reliability Increase from 0.9 to 0.99

Now calculating life adjustment factor for both value of reliability from Weibull parametres


Similarly

Now calculating bearing life for each value

Now using given ball bearing life equation and dividing each other similar to previous problem

Catalog rating increased by factor of 0.61