Answer:
What is true about a point and shoot camera?
- They are often smaller than SLR cameras
Answer:
no
Explanation:
it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls
Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
__
<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
__
<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
Answer:
4 number answer is correct.
is the volume of the sample when the water content is 10%.
<u>Explanation:</u>
Given Data:
First has a natural water content of 25% = 
= 0.25
Shrinkage limit, 
We need to determine the volume of the sample when the water content is 10% (0.10). As we know,
![V \propto[1+e]](https://tex.z-dn.net/?f=V%20%5Cpropto%5B1%2Be%5D)
------> eq 1
The above equation is at 
,
Applying the given values, we get
Shrinkage limit is lowest water content
Applying the given values, we get
Applying the found values in eq 1, we get
