Physics, 19.02.2020 04:55 needhelp9627

Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact is about 2.0 m/s . If this can be reduced to 1.3 m/s or less, the hip will usually not fracture. One way to do this is by wearing elastic hip pads.

A)If a typical pad is 5.0 cm thick and compresses by 2.2 cm during the impact of a fall, what acceleration (in m/s2 ) does the hip undergo to reduce its speed to 1.3 m/s ?

B)What acceleration (in g 's ) does the hip undergo to reduce its speed to 1.3 m/s ?

C)The acceleration you found in part A may seem like a rather large acceleration, but to fully assess its effects on the hip, calculate how long it lasts.

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Answer from: belladaviau1338

(a).The acceleration is 52.5 m/s²

(b). The acceleration is 5.35 g.

(c). The time is 0.038 sec.

Explanation:

Given that,

Initial speed = 2.0 m/s

Distance = 2.2 cm

Final speed = 1.3 m/s

We need to calculate the acceleration

Using equation of motion

v^2=u^2+2as

$a=\dfrac{v^2-u^2}{2s}$

Put the value into the formula

$a=\dfrac{(1.3)^2-(2.0)^2}{2\times2.2\times10^{-2}}$

$a=-52.5\ m/s^2$

Negative sign shows the deceleration

(b). We need to calculate the acceleration in terms g

$a=\dfrac{52.5}{9.8}$

$a=5.35\ g$

The acceleration is 5.35 g.

(c). We need to calculate the time

Using equation of motion

v=u+at

$t=\dfrac{v-u}{a}$

Put the value into the formula

$t=\dfrac{0-2}{-52.5}$

$t=0.038\ sec$

Hence, (a).The acceleration is 52.5 m/s²

(b). The acceleration is 5.35 g.

(c). The time is 0.038 sec.

Answer from: Quest

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 $(-\infty,5]$ .

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typicall...

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