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![a \ student \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\ \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!](https://tex.z-dn.net/?f=%20a%20%5C%20student%20%20%5C%20is%20%5C%20able%20%5C%20to%20%5C%20lift%20%5C%20a%20%5C%20500%28n%29%20%5C%5C%20weight%20%5C%20by%20%5C%20applying%20%5C%20only%20%20%5C%20100%28n%29%20%5C%5C%20%5C%5C%20so%2C%20%5C%20this%20%5C%20info%20%5C%20here%2C%20%5C%20we%20%5C%20simply%20%5C%20divide%20%5C%20by%20%5C%5C%20how%20%5C%20much%20%5C%20this%20%5C%20kid%20%5C%20lifted%2C%20%5C%20by%20%5C%20the%20%5C%20weight%20%5C%20he%2Fshe%20%5C%5C%20%5C%20is%20%5C%20applying.%20%5C%5C%20%5C%5C%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B500%28n%29%2F100%28n%29%7D%7D%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%5C%5C%20and%20%5C%20from%20%5C%20this%2C%5C%20your%20%5C%20answer%20%5C%20would%20%5C%20%5C%5C%20conclude%20%5C%20to%20%5C%20be%20%5C%20%285%29%20%5C%5C%20%5C%5C%20%5Cboxed%7B5%7D%20%5C%20would%20%5C%20be%20%5C%20your%20%5C%20answer%21)
Hope this helps you!
Hope this helps you!
Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the
1 answer:
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Answer:
(A). The work done by friction in crossing the patch is -637.98 J.
(B). The speed of skier is 10.57 m/s.
Explanation:
Given that,
Mass of skier = 62 kg
Speed = 6.5 m/s
Length = 3.50 m
Coefficient kinetic friction = 0.30
Height = 2.5 m
(A) we need to calculate the work done by friction in crossing the patch
Using formula of work done
Put the value into the formula
The work done by friction in crossing the patch is -637.98 J.
(B) we need to calculate the speed of skier
Using conservation of energy
Final potential energy is zero
So,
Put the value into the formula
The speed of skier is 10.57 m/s.
Hence, (A).The work done by friction in crossing the patch is -637.98 J.
(B).The speed of skier is 10.57 m/s.
Answer:
The frequencies are
Explanation:
From the question we are told that
The length of the ear canal is
The speed of sound is assumed to be
Now taking look at a typical ear canal we see that we assume it is a closed pipe
Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as
substituting values
Also the the second harmonic for the pipe (ear canal) is mathematically represented as
substituting values
Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies
