Answer:
A) The updated amplitude = 0.048 m
B) Period T = 0.6 seconds
Explanation: Please refer to the attached documents for the solution.
The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
1 answer:
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1) The wave's period remains constant across different media
2) The wave's velocity varies depending on the medium it travels through
3) As a wave transitions between media, its speed, direction, and wavelength can change, while its frequency stays unchanged
Clarification:
1)
The period of a wave signifies the duration it takes for one full oscillation.
The wave's period is the inverse of its frequency:
where
T denotes the period
f is the frequency
The provided table illustrates that the frequency remains consistent across the three media; hence, the period is unchanged as it solely relies on frequency. We can compute it as we know that
f = 350 Hz
thus the period equals
2)
The velocity of a wave can be derived from the wave equation:
where
f indicates the frequency
is the wavelength
, resulting in a speed of
In the second medium,
, leading to a speed of
In the third medium,
, showing a speed of
As a result, we conclude that the wave's speed varies with the medium.
3)
<pwhen a="" wave="" shifts="" from="" one="" medium="" to="" another="" the="" following="" occurs:="">- The wave's direction alters. Specifically, if the subsequent medium is of greater optical density, the wave bends towards the normal; conversely, it bends away if the second medium is of lesser optical density.
- The wave's speed is affected. The wave decelerates in media with higher optical density and accelerates in those with lower optical density.
- The wave's frequency remains unchanged.
- Ultimately, the wave's wavelength is modified. If moving into a medium of greater optical density, the wavelength decreases, while it increases in one of lower optical density.
Discover more about waves here:
</pwhen></pin>assuming north-south is along the Y-axis and east-west along the X-axis
X = total X-displacement
from the graph, total displacement in the X-direction is computed as
X = 0 - 20 + 60 Cos45 + 0
X = 42.42 - 20
X = 22.42 m
Y = total Y-displacement
from the graph, total displacement in the Y-direction is computed as
Y = 40 + 0 + 60 Sin45 + 50
Y = 90 + 42.42
Y = 132.42 m
To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding
magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m
Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east
Response:
The primary consequence is an increase in induced charge at the nearest points. However, the overall net charge remains zero, meaning it does not influence the flow.
We can utilize Gauss's law to solve this problem
Ф = ∫ e. dA = / ε₀
The flow of the field is directly correlated to the charge within it. Consequently, placing a Gaussian surface beyond the non-conductive spherical shell means the flow will be zero since the sphere’s charge equals the charge induced in the shell, resulting in a net charge of zero. This evaluation shows that the shell effectively obstructs the electric field.
According to Gauss's law, if the sphere is offset, the only effect it generates is an increment in induced charge at the nearest points. Nevertheless, the net charge remains zero, so it does not impact the flow; irrespective of the sphere's position, the total induced charge is consistently equal to the charge on the sphere.