Answer:
5.51 m/s^2
Explanation:
Initial scale reading = 50 kg
assume the greatest scale reading = 78.09 kg
<u>Determine the maximum acceleration for these elevators</u>
At rest the weight is = 50 kg
Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u>
</u>
<u>
</u>At the 10th floor weight = 78.09 kg
Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N
F = change in weight
Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)
50 * a = 275.61
Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2
The conclusion is, medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media.
<h3>
Effect of density on speed of sound</h3>
Sound wave is mechanical wave that requires material medium for its propagation.
A high dense medium, is a medium with closely packed molecules. Since sound wave requires material medium for its propagation, it will travel faster in a high dense medium than a less dense medium.
Thus, the speed of sound increases as the density of the medium increases.
<h3>Speed of sound in the different media</h3>
The conclusion that can be made from the speed of sound in the different media is "Medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media".
Learn more about effect of density on speed of sound here: brainly.com/question/3323620
In the National Grid, the Step Up Transformer increases the amount of voltage that is passing through the grid at any one time. These are used at power stations, as without them, there would not be enough power to transport the electricity around the national grid.
Hope this helps :)
Answer:
By a factor of 1/4.
Explanation:
The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,
in which 
, 
represent the change in momentum and the time taken for that change.
If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by 
, the following manipulation confirms the answer to this question.
![\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5C%5C%5Csmall%20F_1%20%26%3D%5Csmall%20%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B4%5CDelta%20t%7D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%5Cbigg%5B%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B%5CDelta%20t%7D%5Cbigg%5D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7DF%5Cend%7Baligned%7D)
Here 
is the force that was applied to the object previously.
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