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quester [9]
3 years ago
6

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration

less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.37 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 2.1 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.
Physics
1 answer:
miss Akunina [59]3 years ago
7 0

Answer:

1728.42857143 m/s²

0.00155883061577 s

259.264285715 m/s²

0.0103922041051 s

The child will get injured if he/she falls on a hardwood floor

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=2gs\\\Rightarrow v=\sqrt{2gs+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.37+0^2}\\\Rightarrow v=2.69432737432\ m/s

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-2.69432737432^2}{2\times 2.1\times 10^{-3}}\\\Rightarrow a=-1728.42857143\ m/s^2

Magnitude of deceleration is 1728.42857143 m/s²

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-2.69432737432}{-1728.42857143}\\\Rightarrow t=0.00155883061577\ s

Time taken is 0.00155883061577 s

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-2.69432737432^2}{2\times 1.4\times 10^{-2}}\\\Rightarrow a=-259.264285715\ m/s^2

Magnitude of deceleration is 259.264285715 m/s²

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-2.69432737432}{-259.264285715}\\\Rightarrow t=0.0103922041051\ s

Time taken is 0.0103922041051 s

It is likely that the child will get injured if he/she falls on a hardwood floor.

It is less likely that the child will get injured if he/she falls on a carpeted floor.

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In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

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3 years ago
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3 years ago
What is the direction of the magnetic field around a wire carrying a current perpendicularly into this page
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Answer:

Clockwise direction

Explanation:

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The thumb in the direction of current while the finger curl around in the direction of the magnetic field.

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3 0
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5 0
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A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the
Studentka2010 [4]

Answer:

A) the ratio of volumes of the bubble is Vs/Vb= 3.74

B)  would not be safe since Vs/Vb== 3.5 and there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

Explanation:

assuming the gas of the bubble behaves as ideal gas

P * V = n * R * T

where P= absolute pressure, V= volume occupied by the gas, n = number of moles , R = ideal gas constant , T = absolute temperature

if we assume that the mass of the bubble remains constant ( that is, it does not capture other bubbles during ascension of disaggregate into smaller ones and there is no mass transfer into the bubble due to diffusion)

inicial state)  Pb * Vb = n * R * Tb

final  state)  Ps * Vs = n * R * Ts

dividing both equations

(Ps/Pb)(Vs/Vb) = Ts/Tb

therefore

Vs/Vb= (Ts/Tb) (Pb/Ps)

since Tb = 4°C = 277 K and Ts= 23°C = 296 K

Vs/Vb= (Ts/Tb) (Pb/Ps) = (296K/277K)*(3.5 atm/1 atm) = 3.74

B) if the T remains constant Ts=Tb and thus

Vs/Vb= (Ts/Tb) (Pb/Ps)= 1* (Pb/Ps) = 3.5 atm/1 atm = 3.5

it would not be safe since there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

8 0
3 years ago
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