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Inessa05 [86]
2 years ago
12

To ensure that a vehicle crash is inelastic, vehicle safety designers add crumple zones to vehicles. A crumple zone is a part of

a vehicle designed to crumple easily in a crash. Use Newton’s second law to explain why crumple zones reduce the force in a collision. Help me I dont know this
Engineering
1 answer:
spin [16.1K]2 years ago
5 0

Answer:

Explanation:

Answer: With crumple zones at the front and back of most cars, they absorb much of the energy (and force) in a crash by folding in on itself much like an accordion. ... As Newton's second law explains force = Mass x Acceleration this delay reduces the force that drivers and passengers feel in a crash.Sep 30, 2020

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A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi
saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, \tau_{max}, is given by the following formula;

\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )

t_w = 1 cm = 0.01

h = 29 cm = 0.29 m

h_w = 25 cm = 0.25 m

b = 15 cm = 0.15 m

I_c = The centroidal moment of inertia

I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )

I_c = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right )  = 1.2257083\bar 3 \times 10^{-4}

I_c = 1.2257083\bar 3 × 10⁻⁴ m⁴

From which we have;

4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )

Which gives;

W = 11,416.6879 N

\sigma _{b.max} = \dfrac{M_c}{I_c}

\sigma _{b.max} = 1500 N/cm² = 15,000,000 N/m²

M_c = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

M_{max} = \dfrac{W \cdot L}{4}

L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417

L ≈ 0.64417 m ≈ 64.417 cm.

4 0
2 years ago
A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts
lions [1.4K]

Answer:

The coefficient of static friction, μₛ, between the trunk and turntable = 0.32

Explanation:

For this motion of the trunk B,

Initial velocity, v₀ = 0

Tangential Acceleration, a = 0.28 m/s²

Time taken, t = 10s

Using equations of motion,

v = v₀ + at

v = 0 + 0.28 × 10 = 2.8 m/s

Frictional force, Fᵣ = μₛN

μₛ = coefficient of static friction,

N = Normal reaction exerted on the trunk B as a result of its weight = mg

Doing a force balance on the trunk B,

Force keeping the trunk B in circular motion must balance the frictional forces.

Force keeping the trunk B in circular motion, F = mv²/r

Fᵣ = F

μₛN = mv²/r but N = mg

μₛmg = mv²/r

μₛg = v²/r

μₛ = v²/gr

μₛ = 2.8²/(9.8 × 2.5) = 0.32

Hope this helps!!!

3 0
3 years ago
A cylinder contains 480 cm3 of loose dry sand which weighs 820 g. Under a static load of 200 kPa the volume is reduced 1%, and t
goblinko [34]

Answer:

a.

b.

c.

Explanation:

a. void  ratio is provided by the formula: e = \frac{V_{p} }{V_{s}  }

   where , V_{p} = volume of voids

                V_{s} = volume of solid grains

for loose sand, the void space = \frac{480}{480}

                                                   = 1

b. void ratio after static load = 0.1/(480)/ (480)

                                               = 0.1

c. void ratio after vibration = [480- ( 0.1 * 480) ]/ 480

                                             = 0.9

5 0
3 years ago
g (d) Usually, in the case of two finite-duration signals like in parts (a) and (b), the convolution integralmust be evaluated i
Bas_tet [7]

you face is A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

3 0
3 years ago
A seamless pipe carries 2400m³ of steam per hour at a pressure of 1.4N/mm².The velocity of flow is 30m/s.assuming the tensile st
MatroZZZ [7]

Answer:

yessss

Explanation:

7 0
2 years ago
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