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quester [9]
2 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]2 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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A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate
Mandarinka [93]

Answer:

0.456033049

Explanation:

F=\mu N where N=mg hence

F=\mu mg where m is mass of object, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2}, \mu is the coefficient of static friction and F is the applied force.

Making \mu the subject we obtain

\mu=\frac {mg}{N} and substituting m for 38 Kg, g for 9.81 m/s^{2} and 170 N for  F we obtain

\mu=\frac{170} {38*9.81}=0.456033049

Therefore, the coefficient of static friction is 0.456033049

5 0
2 years ago
A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
Ram has power of 550 watt. What does it mean?
WARRIOR [948]
It means you can do 550 Newton Meters of work every second. Power is the rate of doing work, I hope this helps
4 0
2 years ago
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6
Shalnov [3]

Answer:

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

Explanation:

A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.

A) What is the wavefunction y(x,t) for the standing wave that is produced?

B) In which harmonic is the standing wave oscillating?

C) What is the frequency of the fundamental oscillation?

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. lambda=2L/n

when comparing the wave equation with the general wave equation , we get the wavelength to be

2*pi*x/lambda=6.98x

lambda=0.9m

we use the equation

lambda=2L/n

n=number of harmonics

L=length of string

0.9=2(1.35)/n

n=2.7/0.9

n=3

third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

8 0
2 years ago
A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat
weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2

Now we can find the period of the second pendulum at the same location, which is given by

T=2\pi \sqrt{\frac{L}{g}}

where we have

L = 0.36 m (length of the  second pendulum)

g=17.5 m/s^2

Substituting,

T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s

#LearnwithBrainly

8 0
2 years ago
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