<h3><u>Answer;</u></h3>
The period of the wave is <u><em>4 seconds</em></u>
<h3>
<em><u>Explanation;</u></em></h3>
- <em><u>The period of a wave or periodic time is the time taken for one complete oscillation to occur.</u></em> In this case, one complete oscillation occurs when the wave moves from one crest to the next or a trough to the next. <em><u>This takes 4 seconds. Therefore the period is 4 seconds.</u></em>
- <em><u>Frequency on the other hand is the number of oscillations by a wave in one second. Thus, f = 1/T, that is frequency is the reciprocal of periodic time.</u></em>
Pseudoscience because it involves beliefs not facts
Answer:
1.63103 seconds
Explanation:
= Actual period of clock = 1 second
v = Speed at which the observer is moving = 0.79 c
c = Speed of light =
Time dilation is given by
The period is 1.63103 seconds
1) See attached graph
To solve this part of the problem, we have to keep in mind the relationship between current and charge:
where
i is the current
Q is the charge
t is the time
The equation then means that the current is the rate of change of charge over time.
Therefore, if we plot a graph of the charge vs time (as it is done here), the current at any time will be equal to the slope of the charge vs time graph.
Here we have:
- Between t = 0 and t = 2 s, the slope is , so the current is 25 A
- Between t = 2 s and t = 6 s, the slope is , so the current is -25 A
- Between t = 6 s and t = 8 s, the slope is , so the current is 25 A
Plotting on a graph, we find the graph in attachment.
2)
The relationship we have written before
Can be rewritten as
This is valid for a constant current: if the current is not constant, then the total charge is simply equal to the area under the current vs time graph.
Therefore, we need to find the area under the graph.
Here we have a trapezium, where the two bases are
A = 1 ms = 0.001 s
B = 2 ms = 0.002 s
And the height is
h = 10 mA = 0.010 A
So, the area is
So, the charge is .
Answer:
x = 27.3 m
Explanation:
This is a projectile launching exercise, let's start by looking for the time it takes for the rock to reach the height of the window.
Let's use trigonometry to find the velocities of the rock
sin 40 = / v
cos 40 = v₀ₓ / v
v_{oy}= v sin 40
v₀ₓ = v cos 40
v_{oy} = 30 sin 40 = 19.28 m / s
v₀ₓ = v cos 40
v₀ₓ = 30 cos 40 = 22.98 m / s
we look for the time
= v_{oy}^2 - 2 g y
v_{y}^2 = 19.28 2 - 2 9.8 16 = 371.71 - 313.6 = 58.118
v_{y} = 7.623 m / s
we calculate the time
v_{y} = v_{oy} - gt
t = (v_{oy} - v_{y}) / g
t = (19.28 -7.623) / 9.8
t = 1,189 s
since the time is the same for both movements let's use this time to find the horizontal distance
x = v₀ₓ t
x = 22.98 1,189
x = 27.3 m