₦airaland Forum

CondomSir:
Not right.

Using P=hdg, the water exerts maximum pressure on the base when the longest side is placed vertically.

3m is the longest side here.

CondomSir:
Yeah, Lamlad has a similar question.

A tank of dimensions 3mx2mx1.5m is filled with water of density 1000kg/m(cube) and sealed. Neglecting the weight of the tank, calculate the maximum pressure which the tank can maintain on the supporting surface.

Using P=hdg, the water exerts maximum pressure on the base when the longest side is placed vertically.


Also, if P=F/A, a container filled with liquid will exert minimum pressure on the surface when the container has a large area, and maximum pressure when the container has a small area.

CondomSir:
Yeah, Lamlad has a similar question.

A tank of dimensions 3mx2mx1.5m is filled with water of density 1000kg/m(cube) and sealed. Neglecting the weight of the tank, calculate the maximum pressure which the tank can maintain on the supporting surface.

Using P=hdg, the water exerts maximum pressure on the base when the longest side is placed vertically.


Also, if P=F/A, a container filled with liquid will exert minimum pressure on the surface when the container has a large area, and maximum pressure when the container has a small area.

CondomSir:
You can get Lamlad's physics plus the New school physics. Support it with jamb past questions.



You might still want to post more questions sir.

CondomSir:
For minimum pressure, we make use of the two longest surfaces

40cm and 60cm


40cm=0.4m and 60cm=0.6m


Area=0.4x0.6
=0.24m²


F=mg, (using g=9.8m/s²)

F=10x9.8

F=98N


Recall, pressure=FORCE/AREA

Pressure=98/0.24

Pressure=408.333N/m²

CondomSir:
1. Given: v=20m/s, s=50m

Coefficient of dynamic friction(M)=frictional force(Fr)/normal reaction(Fn)..

=ma/mg
=a/g

Recall from kinematics that
v²=u²+ 2as
Since u=0
a=v²/2s
a=20²/2(50)
=400/100
=4m/s²

Therefore
M=4/10
M=0.4

2.Frictional force, Fr= 20N (the horizontal force just sufficient to set the block in motion)
Applied force, F = 40N (when the initial force is doubled)
Net force/Accelerating force,
Fa =F - Fr= 40 - 20 = 20N
Mass, m = 4kg
Acceleration, a = ?
Recall that, net force, Fa = m*a
Therefore, 20 = 4a
a = 20/4
a = 5m/s².

CondomSir:
1. Given: v=20m/s, s=50m

Coefficient of dynamic friction(M)=frictional force(Fr)/normal reaction(Fn)..

=ma/mg
=a/g

Recall from kinematics that
v²=u²+ 2as
Since u=0
a=v²/2s
a=20²/2(50)
=400/100
=4m/s²

Therefore
M=4/10
M=0.4

2.Frictional force, Fr= 20N (the horizontal force just sufficient to set the block in motion)
Applied force, F = 40N (when the initial force is doubled)
Net force/Accelerating force,
Fa =F - Fr= 40 - 20 = 20N
Mass, m = 4kg
Acceleration, a = ?
Recall that, net force, Fa = m*a
Therefore, 20 = 4a
a = 20/4
a = 5m/s².

oldfoolnigger:
yea.

MrHistorian:
SENT.

MrHistorian:
Egbon...You wee nuh kee me with laughter ooooh.

Wah do you mean by "the joke we had earlier"??

We all know how flexible whatsapp group are once they are opened to the public domain.I assure you that there is a very strict administration to this particular one.

Members will not just be added based on "I want to join" plea...It's a group for aspiring scholars only.Personal and Irrelevant chats are prohibited.

Trust me bruh...Nah Confirm be dheez one.

oldfoolnigger:
On behalf of the entire members here,we say a big welcome to u. Ur stays here will surely yield a positive result.





Ur course and ur aggregate score?