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deff fn [24]
3 years ago
6

Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da

mage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a 1000 kg mass vehicle. (a) For a low speed test at 2.5 rnls, compute the energy in the vehicle just prior to impacl If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 em
Physics
1 answer:
valentinak56 [21]3 years ago
3 0

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

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VladimirAG [237]

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that,  F=q*E

The electric field between  the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

3 0
3 years ago
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+v
Arturiano [62]

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

<u>a = 1.72 m/s²</u>

5 0
3 years ago
Air pressure increases as you travel higher above sea level. This is the reason that cabins in commercial airliners require pres
irakobra [83]

The answer is true about the cabins in commercial airliners that require pressurization.

<h3>Why are the cabins of commercial airplanes pressurized?</h3>

Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.

So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.

Learn more about pressure here: brainly.com/question/28012687

#SPJ1

4 0
1 year ago
A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffici
Ivenika [448]

Answer:

Coefficient of friction will be 0.296

Explanation:

We have given initial speed of the stone u = 8 m /sec

It comes to rest so final speed v = 0 m /sec

Distance traveled before coming to rest s = 11 m

According to third equation of motion

v^2=u^2+2as

So 0^2=8^2+2\times a\times 11

a=\frac{-64}{22}=-2.90m/sec^2

Acceleration due to gravity g=9.8m/sec^2

We know that acceleration is given by

a=\mu g

So 2.90=9.8\times \mu \\

\mu =\frac{2.9}{9.8}=0.296

So coefficient of friction will be 0.296

3 0
3 years ago
The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha
gregori [183]

The answers are:

a) Work=125,923.61J

b) Power=1.46watt

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}

Also, we need to find the mass given the density of the blood.

1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg

Now, calculating how much work does the heart do in a day, we have:

Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J

Then, calculating what is the power output and its horsepower, we have:

Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt

Have a nice day!

7 0
3 years ago
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