The net displacement at a point on the string where the pulses cross is 0.2 m.
The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.
A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.
At the post, the pulses are reflected and return along the string without losing any of their amplitude.
Now, let's say the ends are free.
There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.
Now, since A = 0.1 m
Then, 2A = 2(0.1) = 0.2 m
As a result, the net displacement at the string's intersection of two pulses is 0.2 m.
The correct option is (c).
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To calculate for the pressure of the system, we need an equation that would relate the
number of moles (n), pressure (P), and temperature (T) with volume (V). There are a number of equations that would relate these values however most are very complex equations. For
simplification, we assume the gas is an ideal gas. So, we use PV = nRT.<span>
PV = nRT where R is the universal gas
constant
P = nRT / V</span>
<span>P = 3.40 mol ( 0.08205 L-atm / mol-K ) (251 + 273.15 K) / 1.75 L </span>
<span>P = 83.56 atm</span>
<span>
</span>
<span>Therefore, the pressure of the gas at the given conditions of volume and temperature would be 83.56.</span>
Answer:
The maximum height will be 7408.8 meters
Explanation:
final velocity = initial velocity + acceleration × time
final velocity = 0 m/s + 58.8 m/s^2 ×6 s
Final velocity = 352.8 m/s
final velocity ^2 = initial velocity ^2 + 2 × acceleration × displacement
(352.8)^2 = (0)^2 + 2×58.8 ×displacement
Solving for displacement,
height = 1058.4 meters.
After this, the rocket is in free fall, we can use the same equation.
final velocity ^2 = initial velocity ^2 + 2 ×acceleration×displacement
final velocity = 0
0^2 = 352.8^2 + 2×(-9.8)×displacement
displacement = 6350.4 meters
the maximum height will be 7408.8 meters
At the "very top" of the ball's path, there's a tiny instant when the ball
is changing from "going up" to "going down". At that exact tiny instant,
its vertical speed is zero.
You can't go from "rising" to "falling" without passing through "zero vertical
speed", at least for an instant. It makes sense, and it feels right, but that's
not good enough in real Math. There's a big, serious, important formal law
in Calculus that says it. I think Newton may have been the one to prove it,
and it's named for him.
By the way ... it doesn't matter what the football's launch angle was,
or how hard it was kicked, or what its speed was off the punter's toe,
or how high it went, or what color it is, or who it belongs to, or even
whether it's full to the correct regulation air pressure. Its vertical speed
is still zero at the very top of its path, as it's turning around and starting
to fall.
Answer:
The value is 
Explanation:
From the question we are told that
The radius of curvature is 
Generally the radius of curvature is mathematically represented as
Here 
is the focal length hence
=> 
=>
