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Inessa05 [86]
3 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]3 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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Set oWMP = CreateObject("WMPlayer.OCX.7?) Set colCDROMs = oWMP.CdromCollection do if colCDROMs.Count >= 1 then For i = 0 to c
vladimir2022 [97]

Answer:

gg

Explanation:

gg

7 0
3 years ago
Print reading for industry unit 9 review questions
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3 years ago
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Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a
stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0
3 years ago
Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and
lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1]  x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is \frac{1}{x[n] . x[n-1] x[n+1]}

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

                       for x[n] = 2\pi , y[n] = cos(2\pi) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

3 0
2 years ago
In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in
stellarik [79]

Answer:

V = 125.7m/min

Explanation:

Given:

L = 400 mm ≈ 0.4m

D = 150 mm ≈ 0.15m

T = 5 minutes

F = 0.30mm ≈ 0.0003m

To calculate the cutting speed, let's use the formula :

T = \frac{pi* D * L}{V*F}

We are to find the speed, V. Let's make it the subject.

V = \frac{pi* D * L}{F*T}

Substituting values we have:

V = \frac{pi* 0.4 * 0.15}{0.0003*5}

V = 125.68 m/min ≈ 125.7 m/min

Therefore, V = 125.7m/min

7 0
3 years ago
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