Physics, Chemistry and Biology: Post your doubts here!

[Emmris]

Should be lol

 

[Emmris]

Ok, So since the pressure and height is given for the bicycle pump before being pressed (which is 100KPa aka 100 000Pa and Height=45cm) So by using these values you can easily find the density of air by the formula: Pressure= Height*Density*gravitiation force (which is 10). So by reaaranigng the formula we can get Density= Pressure/Height*10 which in this case would be 100 000/10*45
Now after finding the density (222.22 we found above) we will proceed onto the part after the piston has been pushed.
Since the height was 45 before and the piston is pressed 15cm so the height down the barrel will be 45-15=30cm
Now use the pressure formula again and since you already found the density you can now easily find the answer
Pressure=Density*Height*10 So.. Pressure down the barrel (aka applied to the cycle tire)= 222.22*30*10=66666.7Pa (66.7 KPa)

I didn't knew abt the formula pressure=density * height .....

 

[Adeen Atif]

I didn't knew abt the formula pressure=density * height .....

Ohh
Yeah actually there are two
One is Pressure=Force/Area
And the other is particularly for liquids and sometimes gases= Density*Height*10

 

[Emmris]

My teacher didn't gav me dat formula 4 pressure in gas.... I got only Boyle's law ....

 

[Adeen Atif]

My teacher didn't gav me dat formula 4 pressure in gas.... I got only Boyle's law ....

Ohh, but this formula is quite imp she should've given it

 

[Emmris]

Ohh, but this formula is quite imp she should've given it

It's him .... but both my tuition n class teacher didn't gav .... I think dat its not in o level's syllabus

 

[Adeen Atif]

It's him .... but both my tuition n class teacher didn't gav .... I think dat its not in o level's syllabus

Ok...But then why did you get a question related to it?

 

[Emmris]

Ok...But then why did you get a question related to it?

I am my teacher's favourite lol ... so he gav me a "though" question 2 try ... dat u solved easily

 

[Adeen Atif]

I am my teacher's favourite lol ...

That's great

... so he gav me a "though" question 2 try ... dat u solved easily

Happy to help

 

[Emmris]

That's great

Happy to help

Hey! My teacher told me dat the answer is 150kPa

 

[Adeen Atif]

Hey! My teacher told me dat the answer is 150kPa

And since I did't know the exact answer I'm not sure about my answer it could be wrong too. But this was the only way to do so.

I already told you

 

[Emmris]

I already told you

Nah ... it's OK all dat matters is dat u helped me like a true friend

 

[Adeen Atif]

Nah ... it's OK all dat matters is dat u helped me like a true friend

Anytime

 

[studyingrobot457]

Can someone please help and explain me this physics question on pressure .....

-> A bicycle pump of length 45cm contains air with a pressure of 100kPa. In order to force air into the tyre, it is necessary to move the piston 15cm down the barrel. Calculate the pressure inside the tyre assuming the temperature is kept constant.

I would say it is Boyle's law P1V1=P2V2, ik you would be thinking, they only gave us length, but since cross-section is the same, so volume ratio should be also same a length, here is the solution

V1=45
v2=45-15=30
p1=100kpa
p2=?

=(45*100 000)/30=150 000=150 kpa

EZ

*just noticed my post is the 10000th*

 

[Emmris]

I would say it is Boyle's law P1V1=P2V2, ik you would be thinking, they only gave us length, but since cross-section is the same, so volume ratio should be also same a length, here is the solution

V1=45
v2=45-15=30
p1=100kpa
p2=?

=(45*100 000)/30=150 000=150 kpa

EZ

*just noticed my post is the 10000th*

Thank u very very much u helped me a lot

 

[Emmris]

I would say it is Boyle's law P1V1=P2V2, ik you would be thinking, they only gave us length, but since cross-section is the same, so volume ratio should be also same a length, here is the solution

V1=45
v2=45-15=30
p1=100kpa
p2=?

=(45*100 000)/30=150 000=150 kpa

EZ

*just noticed my post is the 10000th*

Ur point was right but my teacher told me dat it was too direct .... the area of the cross section is considered as 'a' .... we multiply the area with the length which will give us the volume.... consider pressure to b found as 'p'

45a * 100 000 = 30a *p

The 'a' is cancelled out n then we r left with :
P = 45 * 100 000 / 30 = 150 000 Pa = 15 kPa

 

[princess of cricket]

Name two substances that pass out of the glomerulus into the renal capsule but are normally absent in the urine of man. what happens to these two substances in the nephron?

Could u guys answer this .

 

[Ayesha Asif333]

Name two substances that pass out of the glomerulus into the renal capsule but are normally absent in the urine of man. what happens to these two substances in the nephron?

Could u guys answer this .

Glucose and salts
They r reabsorbed back by active transport because the body needs it

 

[studyingrobot457]

Ur point was right but my teacher told me dat it was too direct .... the area of the cross section is considered as 'a' .... we multiply the area with the length which will give us the volume.... consider pressure to b found as 'p'

45a * 100 000 = 30a *p

The 'a' is cancelled out n then we r left with :
P = 45 * 100 000 / 30 = 150 000 Pa = 15 kPa

I know that it has to be solved out this way, but after all u eventually are going to get 'a' out by division,so it is a useless step, specially at IGCSE / O-level, no much care is about algebra...

 

[studyingrobot457]

Glucose and salts
They r reabsorbed back by active transport because the body needs it

one thing though, It should not be confused with where they are absorbed, I have seen questions, where they ask so and respond, would be the loop of Henle, which is incorrect because most glucose and salts (ions etc) are absorbed in the first loop/coil after the renal capsule