If something is water-soluble, then that means the substance can be dissolved in water.
So... (water) solubility would be the quality of a substance that allows it to be dissolved in water
I hope this helped! If you have any other questions, just ask, and I hope you have a great day! :)
Answer:
None of the above
It should be position is changing and acceleration is constant.
Explanation:
Since the velocity is changing, this means the object is moving, so the <u>position must also be changing</u>.
Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the <u>acceleration must be constant</u> too.
So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.
So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"
A small vehicle with less mass and with less Kinetic Energy will require less distance to stop than a large vehicle.
Answer:
A. Mrŵ² = ųMg
Ŵ = (ųg/r)^½
B.
Ŵ =[ (g /r)* tan á]^½
Explanation:
T.v.= centrepetal force = mrŵ²
Where m = mass of block,
r = radius
Ŵ = angular momentum
On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.
So there for
Mrŵ² = ųMg
Ŵ = (ųg/r)^½
g = Gravitational pull
ų = coefficient of friction.
B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible
So therefore
N *(sin á) = mrŵ² .....equ 1
Since the car does not slide the net vertical forces must be equal and opposite so therefore
N*(cos á) = mg.....equ 2
Where N is the reaction force of the car on the surface.
Equ 2 becomes N = mg/cos á
Substituting N into equation 1
mg*(sin á /cos á) =mrŵ²
Tan á = rŵ²/g
Ŵ =[ (g /r)* tan á]^½
The same force accelerates a small mass faster than
it accelerates a large mass.
It's easier to get a little red wagon going by pushing it
than it is to get a school bus going by pushing it.
