Answer:
The original length of the specimen is found to be 76.093 mm.
Explanation:
From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:
Initial Volume = Final Volume
πd1²L1/4 = πd2²L2/4
d1²L1 = d2²L2
L1 = d2²L2/d1²
where,
d1 = initial diameter = 19.636 mm
d2 = final diameter = 19.661 mm
L1 = Initial Length = Original Length = ?
L2 = Final Length = 75.9 mm
Therefore, using values:
L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²
<u>L1 = 76.093 mm</u>
Answer:
F(x) = 0 ; x < 0
0.064 ; 0 ≤ x < 1
0.352 ; 1 ≤ x < 2
0.784 ; 2 ≤ x < 3
1 ; x ≥ 3
Explanation:
Each wafer is classified as pass or fail.
The wafers are independent.
Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.
X ~ Bi(n,p)
Where n = 3 and p = 0.6 is the success probability
The probatility function is given by :

Where
is the combinatorial number

Let's calculate f(x) :




For the cumulative distribution function that we are looking for :



The cumulative distribution function for X is :
F(x) = 0 ; x < 0
0.064 ; 0 ≤ x < 1
0.352 ; 1 ≤ x < 2
0.784 ; 2 ≤ x < 3
1 ; x ≥ 3
Answer:
forward voltage triggering
temperature triggering
dv/dt triggering
light triggering
gate triggering
Then turning off;
Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals
Explanation:
hope it helps
Answer:
a. Heat removal rate will increase
b. Heat removal rate will decrease
Explanation:
Given that
One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.
We know that heat transfer in fin given as follows

We know that area

Now when diameter will triples then :





So the new heat transfer will increase by 3 times.
Now when copper rod will replace by aluminium rod :
As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.