Answer:
Outside temperature =88.03°C
Explanation:
Conductivity of air-soil from standard table
K=0.60 W/m-k
To find temperature we need to balance energy
Heat generation=Heat dissipation
Now find the value
We know that for sphere
Given that q=500 W
so
By solving that equation we get
=88.03°C
So outside temperature =88.03°C
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:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V = 
V = Av = 
V = 
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m = 
= 
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 = 
v2 = 
v2 = 6.017 m/s
Answer:
C₁₀ = 6.3 KN
Explanation:
The catalog rating of a bearing can be found by using the following formula:
C₁₀ = F [Ln/L₀n₀]^1/3
where,
C₁₀ = Catalog Rating = ?
F = Design Load = 2.75 KN
L = Design Life = 1800 rev/min
n = No. of Hours Desired = 10000 h
L₀ = Rating Life = 500 rev/min
n₀ = No. of Hours Rated = 3000 h
Therefore,
C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3
C₁₀ = (2.75 KN)(2.289)
<u>C₁₀ = 6.3 KN</u>
