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Mazyrski [523]
2 years ago
13

A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about

head accelerations during impacts on the playing field. The researchers observed 249,613 impacts from 423 football players at nine colleges and high schools and collected collision data from participants in other sports. The accelerations during most head impacts ( > 89% ) in helmeted sports caused head accelerations less than a magnitude of 400 m/s2. However, a total of 11 concussions were diagnosed in players whose impacts caused accelerations between 600 and 1800 m/s2, with most of the 11 over 1000 m/s2.
Use equation 2ax(x−x0)=v2x−v20x to determine the stopping distance. Suppose that the magnitude of the head velocity change was 11 m/s and acceleration was1000 m/s2.
Physics
1 answer:
Olin [163]2 years ago
3 0

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

       =6.05 cm

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The SI units for energy is Joules.

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Muscles make up ore than __ of a person body?<br>a.10<br>b.25<br>c40<br>d.75
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A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1
Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

Height of islandh=1.80\times10^{3}\ m

Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

We need to calculate the distance close to the peak

Using formula of distance

H'=H-h

Put the value into the formula

H'=2397.4-1800

H'=597.4\ m

Hence, The distance close to the peak is 597.4 m.

6 0
3 years ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
2 years ago
In the year 1178. five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon
Lubov Fominskaja [6]

Answer: 1.28 sec

Explanation:

Assuming that the glow following the collision was produced instantaneously, as the light propagates in a straight line from Moon to the Earth at a constant speed, we can get the time traveled by the light applying velocity definition as follows:

V = ∆x / ∆t

Solving for ∆t, we have:

∆t = ∆x/v = ∆x/c = 3.84 108 m / 3.8 108 m/s = 1.28 sec

8 0
2 years ago
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