Answer:
Detection of x-rays from a binary star undergoing mass exchange, where mass of component star can be determined.
If the bulb is in series with something else, then . . .
-- The brightness of the bulb depends on the <em>other</em> device in the circuit.
-- If the other device is designed to use <em>less power</em> than the bulb, then the
other device gets <em>more power</em> than the bulb gets.
-- If the other device is designed to use <em>more power </em>than the bulb, then the
other device gets <em>less power</em> than the bulb gets.
-- If the other device is removed from the circuit, then the bulb doesn't light at all.
This description of the often-screwy behavior of a series circuit may partly explain
why the electric service in your home is not a series circuit.
Answer:
denisity = 52.33 g/c
Explanation:
Density:

We have that m = 785 and that v = 15 c
.

d = 52.33 
Answer:
part a : <em>The dry unit weight is 0.0616 </em>
<em />
part b : <em>The void ratio is 0.77</em>
part c : <em>Degree of Saturation is 0.43</em>
part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>
Explanation:
Part a
Dry Unit Weight
The dry unit weight is given as

Here
is the dry unit weight which is to be calculated- γ is the bulk unit weight given as

- w is the moisture content in percentage, given as 12%
Substituting values

<em>The dry unit weight is 0.0616 </em>
<em />
Part b
Void Ratio
The void ratio is given as

Here
- e is the void ratio which is to be calculated
is the dry unit weight which is calculated in part a
is the water unit weight which is 62.4
or 0.04 
- G is the specific gravity which is given as 2.72
Substituting values

<em>The void ratio is 0.77</em>
Part c
Degree of Saturation
Degree of Saturation is given as

Here
- e is the void ratio which is calculated in part b
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Substituting values

<em>Degree of Saturation is 0.43</em>
Part d
Additional Water needed
For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

Here
is the zero air unit weight which is to be calculated
is the water unit weight which is 62.4
or 0.04 
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction

Now as the volume is known, the the overall weight is given as

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.