Answer and Explanation:
The inertial reference frame is one with constant velocity or non-accelerated frame of reference.
The value of acceleration and velocity change will vary in the two frames and will not be same.
As in case, we observe the acceleration and velocity of a moving train from the platform and the one observed in the train itself will be different.
In case of energy, it is dependent on the frame of reference but the energy change is independent of the frame of reference.
Answer:
Rounded to three significant figures:
(a) 
.
(b) 
.
Explanation:
Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.
Let 
denote a natural number (
.) In a double-split experiment of a monochromatic light:
- A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is:
.
- Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength:
.
<h3 /><h3>Maxima</h3>
The path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where 
in 
.
The path difference increases while moving on the screen away from the center. The first order maximum is at 
where 
.
Similarly, the second order maximum is at 
where 
. For the light in this question, at the second order maximum: 
.
- Central maximum: , such that .
- First maximum: , such that .
- Second maximum: , such that .
<h3>Minima</h3>
The dark fringe closest to the center of the screen is the first minimum. 
at that point.
Add one wavelength to that path difference gives another dark fringe- the second minimum. 
at that point.
- First minimum:
, such that 
. - Second minimum:
, such that .
For the light in this question, at the second order minimum: 
.
Answer:
First answer - (E)
Second answer - (B)
Explanation:
The trade-off here is between TV ADVERTISEMENTS and LETTERS TO POTENTIAL VOTERS. The campaign manager for the candidate who is running for reelection, is trying to decide which of the two factors he should use more of or emphasize. The production function for campaign votes can be simplified as
TVAD + LTPV = CV
This is the production function for campaign votes.
<u>PART A</u>
Describe the production function for campaign votes (in words).
ANSWER: (E)
Television advertisements and (or 'plus') letters to potential voters, produce (or 'equal') campaign votes.
<u>PART B</u>
How might information about this function (such as the shape of the isoquants) help the campaign manager plan strategy?
ANSWER: (B)
If television advertisements and letters to potential voters are perfect complements (complements are goods or actions that 'must' go together or be used together) then the campaign manager should use them in fixed proportions (e.g. in a ratio of 50:50).
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Answer:
<em>Distance= 187.5 m
</em>
<em>Displacement= 12.5 m</em>
Explanation:
<u>Distance And Displacement
</u>
The question doesn't say what the length of the pool is, so we are assuming two things:
The pool is 50 m long.
One lap is one pool length, i.e. 50 m
Grant swims 3 3/4 laps. If we add up the total distance is 50*(3.75)=187.5 m
Grant starts in one end, swims one pool, then swims back one pool. Right now, he has zero displacement
Then he swims another pool and returns only 3/4 of the lap, which means he still needed 1/4*50 m=12.5 to arrive at his starting point. The final displacement is the subtraction of the final position minus the initial position, i.e. 12.5 m - 0 m= 12.5 m
Answer:
Distance= 187.5 m
Displacement= 12.5 m
Answer:
direction : West to East
magnitude : 6.0 * 10^-3
Explanation:
<em>Given data :</em>
Four ( 4 ) monitoring wells
location of wells = corners of 1-ha square
Total piezometric head in each well ;
NE corner = 30.0 m ;
SE corner = 30.0 m;
SW corner = 30.6 m;
NW corner = 30.6 m.
<u>Calculate for the magnitude and direction of the hydraulic gradient </u>
first step ; calculate for area
Area = ( 1 -ha ) ( 10^4 m^2/ha )
= 1 * 10^4 m^2
Distance between the wells = length of side
= √( 1 * 10^4 ) m^2
= 100 m
Direction of Hydraulic gradient is from west to east because the total piezometric head in the west = Total piezometric head in the east
Next determine The magnitude of the hydraulic gradient
= ( 30.6 - 30 ) / 100
= 6.0 * 10^-3
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