According to Newton (2nd law), Force = (mass) x (acceleration)
Substitute what we know : Force = (1,000 kg) x (3 m/s²)
Do the arithmetic: Force = 3,000 kg-m/s² = 3,000 newtons
Sure !
Start with Newton's second law of motion:
Net Force = (mass) x (acceleration) .
This formula is so useful, and so easy, that you really
should memorize it.
Now, watch:
The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?
Net Force = (mass) x (acceleration)
= (5.25 kilograms) x (2.5 m/s²)
Net force = 13.125 newtons .
But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?
==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.
How much ?
(Bella's 15.75 newtons) minus (13.125 that the box feels)
= 2.625 newtons backwards, applied by friction.
Answer:
As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.
Explanation:
Depression in freezing point :
where,
=depression in freezing point =
= freezing point constant
m = molality ( moles per kg of solvent) of the solution
As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:
- If molality of the solution in high the depression in freezing point of the solution will be more.
- If molality of the solution in low the depression in freezing point of teh solution will be lower .
Relative lowering in vapor pressure of the solution is given by :
= Vapor pressure of pure solvent
= Vapor pressure of solution
= Mole fraction of solute
Vapor pressure of the solution is inversely proportional to the mole fraction of solute.
- Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
- lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.
i would say that the child with more linear speed is the cild that is 3 meters away from the center of the merry go round. because the child that is 0.5 meters from the center of the merry go round is less linear because the steering of the merry go round is started from the outer part of the merry go round so it would make more sense that the child that is 3 meters from the center of the merry go round would be more linear in speed.
hope this helps!
