Then I walked into Mr. and Mrs. Frank's bedroom. On the floor, amidst the chaos of papers and books, my eye lit on the little red-orange checkered, cloth-bound diary that Anne had received from her father on her thirteenth birthday. She loves Anne and respects her work.
I hope this helps you.
Answer:
Velocity components
For Acceleration components;
Explanation:
We are given:
We first need to find the radial position r of washer in x-y plane.
Therefore
r = 500 mm
To find length along direction OA we have:
Therefore, the radial and vertical components of velocity will be given as:
Where Q is the angle between OA and vector r.
Therefore,
• 
•
The radial and vertical components of acceleration will be:
Therefore we have:
• 
• 
• 
Note : image is missing, so I attached it
Answer:
D) AND gate.
Explanation:
Given that:
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print
These conditions are:
1. The printer's electronic circuits must be energized.
2. Paper must be loaded and ready to advance.
3. The printer must be "on line" with the microprocessor.
Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.
The objective here is to determine the basic logic gate used in this circuit.
Now;
For NOR gate;
NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.
For NOT gate.
NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.
Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".
Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.
Answer:
option B is correct. Fracture will definitely not occur
Explanation:
The formula for fracture toughness is given by;
K_ic = σY√πa
Where,
σ is the applied stress
Y is the dimensionless parameter
a is the crack length.
Let's make σ the subject
So,
σ = [K_ic/Y√πa]
Plugging in the relevant values;
σ = [50/(1.1√π*(0.5 x 10^(-3))]
σ = 1147 MPa
Thus, the material can withstand a stress of 1147 MPa
So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.
Answer:
the life (N) of the specimen is 46400 cycles
Explanation:
given data
ultimate strength Su = 1600 MPa
stress amplitude σa = 900 MPa
to find out
life (N) of the specimen
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 1600
Se = 800 Mpa
and we know
Se for steel is 700 Mpa for Su ≥ 1400 Mpa
so we take endurance limit Se is = 700 Mpa
and strength of friction f = 0.77 for 232 ksi
because for Se 0.5 Su at 
cycle = (1600 × 0.145 ksi ) = 232
so here coefficient value (a) will be
a = 
a = 
a = 2168.3 Mpa
so
coefficient value (b) will be
a = -
log
b = -log
b = -0.0818
so no of cycle N is
N = 
put here value
N = 
N = 46400
the life (N) of the specimen is 46400 cycles
