Answer:
Given data
w1/w2=6.5/1
Power=5 KW
wp=1800 rpm
angle=14 degrees
Based on above values,the minimum diameter=30 mm
Answer:
design angle ∅ = 4.9968 ≈ 5⁰
Explanation:
First calculate the force Fac :
Fac = 
= 
= 708.72 Ib
using the sine law to determine the design angle

hence ∅ = 
=
= 4.9968 ≈ 5⁰
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:
v₀ = 2,562 m / s = 9.2 km/h
Explanation:
To solve this problem let's use Newton's second law
F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v
F dx = m v dv
We replace and integrate
-β ∫ x³ dx = m ∫ v dv
β x⁴/ 4 = m v² / 2
We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max
-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)
x_max⁴ = 2 m /β v₀²
Let's look for the speed that the train can have for maximum compression
x_max = 20 cm = 0.20 m
v₀ =√(β/2m) x_max²
Let's calculate
v₀ = √(640 106/2 7.8 104) 0.20²
v₀ = 64.05 0.04
v₀ = 2,562 m / s
v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)
v₀ = 9.2 km / h
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.