Ask question

Ask question

All categories

2 years 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?

1 answer:

rusak2 [61]2 years ago

5 0

Answer:

change in KE = -12.95 Btu

Explanation:

given data

mass = 3000-lbm

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

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

solution

as here for crumple zone to protected zone it will absorb kinetic energy

so change in KE is related here initial and final velocity that is

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

put here value and we get

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

You might be interested in

Car A rounds a curve of 150‐m radius at a constant speed of 54 km/h. At the instant represented, car B is moving at 81 km/h but

Snezhnost [94]

Answer:

Velocity of Afrom B=21m/s

Acceleration of A from B=1.68m/s°2

Explanation:

Given

Radius r=150m

Velocity of a Va= 54km/hr

Va=54*1000/3600=15m/s

Velocity of b Vb=82km/hr

VB=81*1000/3600=22.5mls

The velocity of Car A as observed from B is VBA

VB= VA+VBA

Resolving the vector into X and Y components

For X component= 15cos60=7.5m/s

Y component=22 5sin60=19.48m/s

VBA= √(X^2+Y^2)

VBA= ✓(7.5^2+19.48^2)=21m/s

For acceleration of A observed from B

A=VA^2/r= 15^2/150=1.5m/s

Resolving into Xcomponent=1.5cos60=0.75m/s

Y component=3cos60=1.5

Acceleration BA=√(0.75^2+1.5^2)

1.68m/s

4 0

2 years ago

Read 2 more answers

if one sprinter runs the 400.0 m in 58 seconds and another can run the same distance in 60.0 seconds, by how much distance will

SCORPION-xisa [38]

v = 400m/58s = 6.9m/s

v = 400m/60s = 6.7m/s

6.9m/s - 6.7m/s = .2m/s difference

60sec - 58 sec = 2 sec

v =Δx/t

Δx = vt

Δx = (.2m/s)(2sec)

Δx = .4m

therefore, the answer is .4m

4 0

2 years ago

Read 2 more answers

The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1

antoniya [11.8K]

Answer:

Explanation:

Constant pressure molar heat capacity Cp = 29.125 J /K.mol

If Cv be constant volume molar heat capacity

Cp - Cv = R

Cv = Cp - R

= 29.125 - 8.314 J

= 20.811 J

change in internal energy = n x Cv x Δ T

n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature

Putting the values

= 20 x 20.811 x 15

= 6243.3 J.

3 0

2 years ago

A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s

BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

$Sf^2=So^2+(2*a*x)$

$Sf=So+(a*t)$

We are going to put our data:

$(15m/s)^2=So^2+(2*a*60m)$

$15m/s=So+(a*6s)$

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

$15m/s-(a*6s)=So$

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

$(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}$

$-120m*a=-180m*a+36s^{2}*a^{2}$

$0=120m*a-180m*a+36s^{2}*a^{2}$

$0=-60m*a+36s^{2}*a^{2}$

$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

$1.67m/s^{2}=a2$

If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

$Sf=So+(a*t)$

$15m/s=So+1.67m/s^2*6s$

$15m/s-(1.67m/s^2*6s)=So$

$4.98m/s=So$

For part c:

We are going to use:

$Sf^2=So^2+(2*a*x)$

$(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)$

$\frac{24.80m^2/s^2}{3.34m/s^2}=x$

$7.42m=x$

5 0

2 years ago

The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel

ANTONII [103]

Answer:

B

Explanation:

7 0

2 years ago

Read 2 more answers