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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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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height
lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u>  m.

When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u>  m.

When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is  <u>0.87</u> m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is  <u>1.25</u> m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is  <u>1.86</u> m.

Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!

6 0
2 years ago
Read 2 more answers
A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas
vodka [1.7K]

Answer:

v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

F=qE

F_p=F_e

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

\frac{m_ev_e^2}{2d}=qE

we can equals these expressions for both proton and electron, because the forces qE are the same:

\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

4 0
2 years ago
Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and
scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r  were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U =  -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

7 0
2 years ago
A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the
Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

d^{2} =\sqrt{ 2.05*10^-7}

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0
2 years ago
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