Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
Given:
diameter of sphere, d = 6 inches
radius of sphere, r =
= 3 inches
density,
= 493 lbm/ 
S.G = 1.0027
g = 9.8 m/
= 386.22 inch/ 
Solution:
Using the formula for terminal velocity,
=
(1)

where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A = 
Volume of sphere, V = 
Using the above formulae in eqn (1):
= 
=
= 
Therefore, terminal velcity is given by:
=
inch/sec
Answer:
Since the farmer wants to buy a 10 kg bag of fertilizer, he should buy it from merchant A. However, Merchant A and B are selling at the same price for a unit value. In other words, Both Merchant A and B are selling 1kg of dry fertilizer for $1.
Explanation:
Which merchant has the better deal means which merchant offers the farmer a better deal.
For Merchant A, 10 kg bag = $10
meaning it contains a real 10 kg bag of dry fertilizer which the farmer can use without losing any Kg to drying.
While for Merchant B, 10 kg bag = $8
where the 10kg = 80% dry fertilizer + 20% water content
But the farmer can only use the solid constituents of the bag which means,
Merchant B is giving 80/100 x 10Kg of dry fertilizer for $8
That is, 8kg for $8
Since the farmer wants to buy a 10 kg bag of fertilizer, he should buy it from merchant A. However, Merchant A and B are selling at the same price for a unit value. In other words, Both Merchant A and B are selling 1kg of dry fertilizer for $1.
Answer:
Timing Diagrams 15 pts. A 10 MHz clock that generates a 0 to 5V pulse train with a 30% duty cycle is connected to input X of a two input OR gate that has a 20nS propagation delay. The clock also goes to an inverter with a 10 ns propagation delay. The output of the inverter goes to the Y input of the OR gate. a) Draw the circuit. 2 pts. b) Plot the output of the clock for two cycles. Show times and voltages. 5 pts. c) On the same page as part (b) plot the output of the inverter. Show times and voltages. 3 pts. d) On the same page as parts (b & c) plot the output of the OR gate. Show times and voltages. 5 pts.