Answer:
Scientific Notation: 3.45 x 10^5
E Notation: 3.45e5
Answer:
8.89288275 m/s
Explanation:
F = Tension = 54 N
= Linear density of string = 5.2 g/m
A = Amplitude = 2.5 cm
Wave velocity is given by

Frequency is given by

Angular frequency is given by

Maximum velocity of a particle is given by

The maximum velocity of a particle on the string is 8.89288275 m/s
A concave mirror is opaque whereas a concave lens is transparent. ... Difference between concave mirror and concave lens is A concave mirror is opaque whereas a concave lens is transparent. A concave mirror causes reflection of light whereas a concave lens causes refraction of light.
The heat <span>Q(in)</span> supplied to the system in one stage of the cycle, minus the heat <span>Q(out)</span> removed from it in another stage of the cycle; plus the work added to the system <span>W(in)</span> equals the work that leaves the system <span>W(<span>out)</span></span>
Answer:
150m
Explanation:
The relation of speed/time and distance/time is a derivative/integral one, as in speed is the derivative of distance (the faster you go, the faster the distance changes, duh!).
So we need to compute the integral of speed over time from 0.0s to 5.0s.
The easiest way here is to compute the area under the line (it's going to be faster than computing the acceleration and using a formula of distance based on acceleration).
The area under the line is a trapezoid with "height" 5s, and the bases 10m/s and 50m/s. Using the trapezoid area formula of h*(a + b)/2
distance = 5s * (10m/s + 50m/s) / 2 = 5s * 60m/s / 2 = 5s * 30m/s = 150m
Alternatively, we can use the acceleration formula:
a = (50m/s - 10m/s)/5s = 40m/s / 5s = 8m/s^2
distance = v0 * t + a * t^2 / 2 = 10m/s * 5s + 8m/s^2 * (5s)^2 / 2 = 50m + 8m * 25 / 2 = 50m + 100m = 150m.