What do radio waves and micowaves have in common
1 answer:
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The formula for the period of wave is: wave period is equals to 1 over the frequency.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>
</h2><h2>_____________________________________</h2><h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
<h2>SOLUTION:</h2>According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.
To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,
Base is the x-axis of the graph which is Position i.e. 6m
Height is the y-axis of the graph which is Force i.e. 30N
So,
W = 90 J
The work done is 90 J.
According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.