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erastovalidia
2 months ago
15

Vehicle crumple zones are designed to absorb energy during an impact by deforming to reduce energy transfer to the occupants. Ho

w much kinetic energy, in Btu, must a crumple zone absorb to fully protect occupants in a 3000-lbm vehicle that suddenly decelerates from 10 mph to 0 mph?
Physics
1 answer:
kicyunya [3.2K]2 months ago
5 0

Answer:

change in KE = -12.95 Btu

Explanation:

provided data

mass = 3000-lbm

initial velocity of vehicle vi = 10 mph = 14ft/s

final velocity of vehicle vf = 0 mph = 0 ft/s

solution

the crumple zone is designed to absorb kinetic energy upon impact

thus the change in KE is related to the initial and final speeds, expressed as:

change in KE = 0.5 × m × (vf² - vi² ).................1

Substituting in the parameters yields:

change in KE = 0.5 × 3000 × (0² - 14.7² )

change in KE = -324135 × \frac{1lbf}{32.174\ lb.ft/s^2} * \frac{1Btu}{778.17 ft.lbf}  

change in KE = -12.95 Btu

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A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer
Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk m=12.5 kg

diameter of disc R=0.23 m

weight of ring m_r=7 kg

Force F=9.7 N

N=180 rpm

\omega =\frac{2\pi N}{60}

\omega =6\pi rad/s

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2

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At this point, Torque is T=F\times R=I\cdot \alpha

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\omega _f=0 in this scenario

0=6\pi -3.18\times t

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7 0
1 month ago
A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the
Yuliya22 [3333]

The answer is letter d, 8.3.

 

Here’s a solution for the given problem:

 

We have:

B = 6i - 8j 


Let A be unknown; we'll denote A as = mi + nj 



The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

 

and we also know the magnitude of A+B is equivalent to the magnitude of A,

 

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 



Using vector addition, A+B becomes (m+6)i + (n-8)j.

 

Since we know A+B = Ki + 0j, we can establish that: 

m + 6 = K 


n - 8 = 0, which gives n=8. 

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2 


= m^2 + 12m + 36 = m^2 + 64 


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5 0
2 months ago
Read 2 more answers
An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10
inna [3103]

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

m = \dfrac{15}{32.174}\ slug

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

W = \dfrac{1}{2} m (v_f^2-v_i^2)

7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2

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The acceleration of the object is therefore

  a = \dfrac{v_f-v_o}{t}

  a = \dfrac{182.75-0}{10}

         a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²

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