Answer:
<em>10amps</em><em> </em><em>÷</em><em>6</em><em> </em><em>ohms</em><em> </em><em>=</em><em> </em><em>5</em><em>/</em><em>3</em><em> </em><em>(</em><em>1</em><em>.</em><em>6</em><em>)</em>
Answer:
Graph C
Explanation:
This is the answer because it is the only one that shows the vehicle accelerate to a constant speed before stopping and slowing down.
No net force and the photo shows
Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?
B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?
Answer:
The frequency change percentage is 9.94%
Explanation:
The frequency shift can be calculated as follows.

= 
=9.95 GHz
So the frequency change seen by the detector is 9.95 - 9.05
% difference
= 9.94%
Answer:
It's due to the distance from either ends of strings origin...
Explanation:
As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.