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Yuri [45]
3 years ago
8

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 2.0 cm during the impact of a fall, what constant acceleration (in m/s2 and in g’s) does the hip undergo to reduce its speed from 2.0 m/s to 1.3 m/s? (b) The acceleration you found in part (a) may seem rather large, but to assess its effects on the hip, calculate how long it lasts.
Physics
1 answer:
Westkost [7]3 years ago
7 0

Answer:

-5.8868501529 m/s² or -5.8868501529g

0.118909090909 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{1.3^2-2^2}{2\times 0.02}\\\Rightarrow a=-5.8868501529\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{-57.75}{9.81}\\\Rightarrow a=-5.8868501529g

The acceleration is -5.8868501529 m/s² or -5.8868501529g

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{1.3-2}{-5.8868501529}\\\Rightarrow t=0.118909090909\ s

The time taken is 0.118909090909 s

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6 0
3 years ago
The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.
Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let \rho_{m} be the density of metal and \rho_{w} be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

W =0.10400\ N

mg=0.10400

\rho V g=0.10400

Vg=\dfrac{0.10400}{\rho_{m}}.....(I)

The weight of metal in water is

Using buoyancy force

F_{b}=0.10400-0.08400

F_{b}=0.02\ N

We know that,

F_{b}=\rho_{w} V g....(I)

Put the value of F_{b} in equation (I)

\rho_{w} Vg=0.02

Put the value of Vg in equation (II)

\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02

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6 0
3 years ago
A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

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