D. 128.25 because force=mass x acceleration
The period of the wave is the reciprocal of its frequency.
1 / (5 per second) = 0.2 second .
The wavelength is irrelevant to the period. But since you
gave it to us, we can also calculate the speed of the wave.
Wave speed = (frequency) x (wavelength)
= (5 per second) x (1cm) = 5 cm per second
Answer:
B. 22,22,23,23,22,22,23
Explanation:
The standard deviation is a measure of dispersion or variability of a data set. In order to determine the data set that has the smallest standard deviation, we shall investigate on the ranges of the data sets given. The range of a data set is simply the difference between the maximum and minimum values in a data set. A data set that has a smaller range also has a smaller standard deviation.
From the alternatives given, the data set given by alternative B has the smallest range and consequently the smallest standard deviation.
The maximum value is 23 while the minimum is 22. The range is 1.
Answer:
All the given options will result in an induced emf in the loop.
Explanation:
The induced emf in a conductor is directly proportional to the rate of change of flux.

where;
A is the area of the loop
B is the strength of the magnetic field
θ is the angle between the loop and the magnetic field
<em>Considering option </em><em>A</em>, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.
<em>Considering option </em><em>B</em>, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.
Option C has a similar effect with option A, thus both will result in an induced emf.
Finally, <em>considering option</em> D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will<em> </em>change the angle<em> </em>between the loop and the magnetic field. This effect will also result in an induced emf.
Therefore, all the given options will result in an induced emf in the loop.
Kinetic Energy is calculated by KE= 1/2mv2