Physics, 05.02.2020 09:50, jeremymcdonaldorlh0s

If a 70-kg swimmer pushes off a pool wall with a force of 250 n, at what rate will the swimmer accelerate from the wall

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Whenever we do force and acceleration problems, we normally
ignore air resistance. The problem would be too complicated if
we tried to include it, and we don't have enough information to
account for air resistance anyway.

But I think we're going too far if we ignore water resistance for this
question. The equation for Newton's second law ... F = M · a ...
is talking about the NET force on an object. I think we can all
agree that the 250 N of force with which the swimmer pushes
away from the wall is NOT the net force on the swimmer.

As soon as the swimmer starts moving through the water, the water
resists the motion with a bunch of force. The magnitude of the force
due to water resistance depends the swimmer's size, shape, and posture,
plus what kind of swimsuit s/he's wearing and how much hair s/he has.

The question doesn't give us any of that information, and even if we
had it, it would take a team of hydraulic engineers to use it properly.
It's certainly way over MY head.

So I'm going to say that this question can't be answered as it's written.

IF the 250 N were the total, NET of ALL forces acting on the swimmer,
then

Acceleration = (force) / (mass) = (250 N) / (70 kg)

= 3.57 m/s² .

But you and I both know that it isn't.

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

answered: Guest

The relationship of both of them

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If a 70-kg swimmer pushes off a pool wall with a force of 250 n, at what rate will the swimmer accel...