Solve Your Confusion: Explaining Airbags, Braking Force, Circuit Breakers & More

[reree17]

Seriously confused, please help!

Hi there. These are just a few questions I just can’t get my head around. I don’t understand HOW:

a) HOW an air bag reduces injury exactly – Question says ‘Explain how an air bag reduces the risk of head injury in an accident’ – I know that the air bag inflates so it blocks the driver’s access to the steering wheel and thus reducing the risk of head injury, but how can I explain this more scientifically?

b) Calculate the average braking force acting on a car of 800 kg to bring it to rest if it was traveling at 20 m/s – I used the equation F= M X A (Mass x Acceleration). So speed is measured in m/s right? And acceleration is measured in m/s squared? So would I do 800 x 20 squared = 320,000 N or just leave it at 20 m/s?

c) [URL]http://postimage.org/image/1spa51pic/[/URL] – Sorry the picture is a bit blurry, in case you can’t read it it says ‘The circuit below is used to investigate how the resistance of a lamp changes. Explain how component X allows a set of results to be obtained. – I wrote that Component X is a variable resistor and therefore varies the current. The higher the current the lower the resistance and the lower the current the higher the resistance – Is this the right answer? How can it be improved?

d) [URL]http://postimage.org/image/1mxtsq344/[/URL] – Explain how the graph shows that the resistance of the lamp increases as the voltage increases – Following Ohm’s Law, The higher the current, the higher the voltage right? But the higher the current, the lower the resistance? And the higher the voltage the lower the resistance? So why does the resistance increase as the voltage increases?

e) Explain how the miniature circuit breaker protects against fire – I know that the electromagnets in the mcb become strong enough to separate a pair of contacts, but how exactly does separating a pair of contacts protect against a fire?

f) Explain how the residual current device protects the user – I know that the rcd detects a difference in the currents in the live and neutral wires but how does this protect the user? Does it break the circuit or trip the circuit? And what exactly does tripping a circuit mean?

These are just few problems from practice papers I found difficult, I’m trying to understand them to help prepare for my exam. Thanks so so so so much for answering in advance!

 

[Quinzio]

a) HOW an air bag reduces injury exactly – Question says ‘Explain how an air bag reduces the risk of head injury in an accident’ – I know that the air bag inflates so it blocks the driver’s access to the steering wheel and thus reducing the risk of head injury, but how can I explain this more scientifically?

What's your idea ?
Why banging your head against a wall hurts, but it doesn't hurt against a cushion ?

 

[reree17]

What's your idea ?
Why banging your head against a wall hurts, but it doesn't hurt against a cushion ?

Aahh, so it's because the steering wheel is a hard object and therefore can cause injuries such as bone fractures?

 

[Quinzio]

Aahh, so it's because the steering wheel is a hard object and therefore can cause injuries such as bone fractures?

Ok , this is common sense, no one is impressed.
Let's model the head as a cube, so it's more easy to deal with.
Now model the steering wheel into something, and "model" the airbag. Note the quotes ("").
Why the airbag doesn't break the facial bones ?

 

[Ray Vickson]

Aahh, so it's because the steering wheel is a hard object and therefore can cause injuries such as bone fractures?

Besides a skull fracture you need also to worry about internal brain injury, caused by the brain butting hard up against a suddenly-stopped skull. This is the type of injury a boxer can suffer, even though he was struck by a padded boxing glove. The air bag reduces the deceleration (that is, spreads it over a longer period), so reduces the severity of both external and internal skull injuries.

RGV

 

[reree17]

Ok , this is common sense, no one is impressed.
Let's model the head as a cube, so it's more easy to deal with.
Now model the steering wheel into something, and "model" the airbag. Note the quotes ("").
Why the airbag doesn't break the facial bones ?

It's soft? But I thought the airbag on impact can cause injuries itself? Although it prevents the driver from coming into contact with the steering wheel? :S

 

[Disconnected]

It's soft? But I thought the airbag on impact can cause injuries itself? Although it prevents the driver from coming into contact with the steering wheel? :S

Do you know the equation ft=mv-mu? Or f=ma?
If your head comes to rest more slowly, ie has a lower acceleration, then the force exerted on your face is lower, hence less likely to jack you up.

 

[reree17]

Do you know the equation ft=mv-mu? Or f=ma?
If your head comes to rest more slowly, ie has a lower acceleration, then the force exerted on your face is lower, hence less likely to jack you up.

Ah it makes sense now! Thanks all! Do you have any idea about the other questions? I still can't work them out

 

[Disconnected]

Calculate the average braking force acting on a car of 800 kg to bring it to rest if it was traveling at 20 m/s – I used the equation F= M X A (Mass x Acceleration). So speed is measured in m/s right? And acceleration is measured in m/s squared? So would I do 800 x 20 squared = 320,000 N or just leave it at 20 m/s?

Uh, could you clarify what you did here? I can't quite follow it.

Are you just squaring the speed? Because that isn't the same as acceleration. The units of speed squared is m sqaured/s squared. not m/s squared!

 

[reree17]

Uh, could you clarify what you did here? I can't quite follow it.

Are you just squaring the speed? Because that isn't the same as acceleration. The units of speed squared is m sqaured/s squared. not m/s squared!

I used the equation Force = Mass x Acceleration. The mass of the car is 800 kg, but the units of acceleration are m/s squared? And the speed of the car is 20 m/s, the units are not m/s squared. So would I leave it at 20 m/s or convert it to suit the acceleration units?

Oh so going by that logic I just leave it at 20 m/s?

 

[cepheid]

I used the equation Force = Mass x Acceleration. The mass of the car is 800 kg, but the units of acceleration are m/s squared? And the speed of the car is 20 m/s, the units are not m/s squared. So would I leave it at 20 m/s or convert it to suit the acceleration units?

Oh so going by that logic I just leave it at 20 m/s?

Acceleration is something totally different from speed. You haven't been given the acceleration. You have to compute it (hint: what is the definition of acceleration?). You know that the initial speed is 20 m/s, and that the final speed is 0 m/s. But in order to compute the acceleration, you also need to know how long it took for that change in speed to occur (i.e. what was the time interval over which the object came to rest?). If the problem doesn't give you that, then you can't answer it, because you have no way to figure out what the acceleration was.

 

[reree17]

Acceleration is something totally different from speed. You haven't been given the acceleration. You have to compute it (hint: what is the definition of acceleration?). You know that the initial speed is 20 m/s, and that the final speed is 0 m/s. But in order to compute the acceleration, you also need to know how long it took for that change in speed to occur (i.e. what was the time interval over which the object came to rest?). If the problem doesn't give you that, then you can't answer it, because you don't know what the acceleration was.

Acceleration is the rate of the change of velocity over time.
But the question doesn't state the time. Understandably because it is asking for the braking force. Aah ok. Thanks!

How do I work out the braking force given the following information:
Calculate the average braking force acting on a car of 800 kg to bring it to rest if it was traveling at 20 m/s.
The question provides a table which at 20 m/s the braking distance is 30m and 160,000 J.

?

 

[Disconnected]

Acceleration is the rate of the change of velocity over time.
But the question doesn't state the time. Understandably because it is asking for the braking force. Aah ok. Thanks!

How do I work out the braking force given the following information:
Calculate the average braking force acting on a car of 800 kg to bring it to rest if it was traveling at 20 m/s.
The question provides a table which at 20 m/s the braking distance is 30m and 160,000 J.

?

Ahh, there you go! If the distance is 30m, then you can use
v^2=u^2+2as.

Here v, the final velocity, is zero (because it stopped!), u, the initial velocity, is 20 m/s. s, the distance, is 30m

Seems like all you need is a wee bit of rearranging!

 

[reree17]

Ahh, there you go! If the distance is 30m, then you can use
v^2=u^2+2as.

Here v, the final velocity, is zero (because it stopped!), u, the initial velocity, is 20 m/s. s, the distance, is 30m

Seems like all you need is a wee bit of rearranging!

That equation is unfortunately not on our syllabus :( We are only allowed to use Acceleration = Change in velocity / time. How would you implement into this equation? (or f=mxa)

 

[cepheid]

The question provides a table which at 20 m/s the braking distance is 30m and 160,000 J.

?

Yeah, that was kind of an important piece of information that you ought to have included from the start! That's why the homework help posting template asks you for the full problem statement.

Anyway, I should point out that since you're given the work done (160,000 J) by the braking force, you don't even need to use the formula suggested by Disconnected. You can just use work = force * distance (but you get the same answer using either method).

EDIT: To use the formula suggested by Disconnected, just plug in the relevant quantities!

 

[reree17]

Yeah, that was kind of an important piece of information! That's why the homework help posting template asks you for the full problem statement.

Anyway, I should point out that since you're given the work done (160,000) by the braking force, you don't even need to use the formula suggested by Disconnected. You can just use work = force * distance (but you get the same answer using either method).

EDIT: To use the formula suggested by Disconnected, just plug in the relevant quantities!

Oh yes! So I was in fact using the wrong equation. All I needed to do was re-arrange the work = force * distance equation to give force = work / distance. Makes sense now, thanks so much!