Answer:
The displacement of the train in this time period is 2,616.86 m.
Explanation:
A Uniformly Varied Rectilinear Motion is Rectilinear because the mobile moves in a straight line, Uniformly because of there is a magnitude that remains constant (in this case the acceleration) and Varied because the speed varies, the final speed being different from the initial one.
In other words, a motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time (the speed is constant and the acceleration is variable).
An independent equation of useful time in this type of movement is:
<em>Expression A</em>
where:
- vf = final velocity
- vi = initial velocity
- a = acceleration
- d = distance
The equation of velocity as a function of time in this type of movement is:
vf=vi + a*t
So the velocity can be calculated as: 
In this case:
- vf=42.4 m/s
- vi=27.5 m/s
- t=75 s
Replacing in the definition of acceleration: 
a=0.199 m/s²
Now, replacing in expression A:
Solving:
d= 2,616.86 m
<u><em>The displacement of the train in this time period is 2,616.86 m.</em></u>
Answer is C: Ability to see three-dimensional images of the surfaces of object
Explanation:
To enable the technician see fractures and broken particles in a better resolution as the SEM sees the peaks and valley of the structure.
The actual answer is 165 miles, but using significant figure rules the answer is 200. This is because the sig fig rules are as follows ...
<span>1. Non-zero digits are always significant.
2. Any zeros between two significant digits are significant.
<span>3. A final zero or trailing zeros in the decimal portion ONLY are significant.
</span></span>
So the zeroes in a number like 20 or 23,000 are NOT significant. When you add numbers you must find the addend with the lowest amount of significant figures and round the answer to that. In this case most of the addends only have one sig fig, so you round 165 to 200 to make it only have one sig fig.
Given:
f1 = 20 Hz
f2 = 20000 Hz
speed of sound at 20 degrees celcius = 343 m/s
Solution:
for f1 = 20 Hz,
Using the equation:
lambda = speed of sound / f1 = 343 / 20 = 17.15 m
For f2:
lambda = speed of sound / f2 = 343 / 20000 = 0.01775 m
Therefore the wavelength range of audible sound in air would be 17.15 m to 0.01775 m.
Answer:
thermal
Explanation:
sorry if you get it wrong
