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Svetradugi [14.3K]
2 years ago
14

Upon impact, bicycle helmets compress, thus lowering the potentially dangerous acceleration experienced by the head. A new kind

of helmet uses an airbag that deploys from a pouch worn around the rider's neck. In tests, a headform wearing the inflated airbag is dropped from rest onto a rigid platform; the speed just before impact is 6.00 m/s. Upon impact, the bag compresses its full 12.0 thickness, slowing the headform to rest.
Required:
Determine the acceleration of the head during this event, assuming it moves the entire 12.0 cm.
Physics
1 answer:
Dimas [21]2 years ago
6 0

Answer:

acceleration = -15.3g

Explanation:

given data

speed = 6.00 m/s.

thickness = 12

moves the entire = 12.0 cm

solution

we will use here equation that is

v² - u²  = 2 × a × s    ........................1

here v = 0 is the final velocity and u = 6.0 m/s is initial velocity and s= 0.12 m is the distance covered and a is the acceleration

so we put here value and get acceleration

a = \frac{v^2-u^2}{2s}

a = \frac{0^2-6^2}{2\times 0.12}

a = -150 m/s² ( negative sign means it is a deceleration )

and

acceleration in units of g  

a = \frac{-150}{9.8}

a = -15.3 g

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nalin [4]

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

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There must be a force.

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So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

3 0
2 years ago
A ray of yellow light ( f = 5.09 × 1014 hz) travels at a speed of 2.04 × 108 meters per second in
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Velocity = fλ

where f is frequency in Hz, and λ is wavelength in meters.

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6 0
2 years ago
Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =
Semenov [28]
For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
r_1+r_2=L km
The second equation is going to the total force acting on the charge q3:
F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}
Let's solve this for r_1^2:
0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0
Now we have a quadratic equation with following parameter:
a=2\\ b=2L\\ c=-L^2
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We need a positive solution. When we plug in all the numbers we get:
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6 0
2 years ago
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Explanation:

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Answer:

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