Well, first of all, if the ramps are perfectly frictionless, then the ball
won't roll ... it'll slide. Kind of like a bowling ball does at the beginning
of a shiny, highly-waxed lane. You need friction against the ball to roll it.
But in any case, no matter what the slopes of the ramps are, even
if they have different slopes, the ball will rise on the second ramp
to the same height from which it was released on the first ramp.
Answer:
It's because it tips over the threshold from nucleate boiling, which we can see, to convection boiling, which we can't. ... Even if the steam stayed in the pot, it would still stop boiling when you removed the heat. The steam and water in a liquid/vapour mixture are at the same temperature (100ºC).
Explanation:
Answer:
The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
Explanation:
Given;
the speed of the slower runner, u₁ = 11.8 km/hr
the speed of the fastest runner, u₂ = 15 km/hr
distance, d = 8 km
The time when the fastest runner finishes the race is given by;
The distance covered by the slower runner at this time is given by;
d₁ = u₁ x 0.533 hr
d₁ = 11.8 km/hr x 0.533 hr
d₁ = 6.29 km
Additional distance (x) the slower runner need to finish is given by;
6.29 km + x = 8km
x = 8 k m - 6.29 km
x = 1.71 km
Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
So first make a force diagram. I would label forward forces + and backward forces -. Essentially, the drag force is equal to the Ft (force of tension) + Ff (force of friction on snowmobile - driver system). The force of Friction is equal to mu * Fn. We can find mu through the force of friction acting on the sled. 120 N = mu * Fn (equal to m * g of sled). mu of the Ice is equal to 0.167. So, 540 N = Ft + 0.167 * 4500 N. Ft = -211.5 N. <u>Ft is acting in the backwards direction at a magnitude of 211.5 N</u>
Answer:
The average velocity is 2 m/s.
Explanation:
The velocity of the particle is the time derivative of its position 
![$v =\frac{dx(t)}{dt} = \frac{d}{dt}[t^2+2] $](https://tex.z-dn.net/?f=%24v%20%3D%5Cfrac%7Bdx%28t%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%5Bt%5E2%2B2%5D%20%24)
Now the average from 
and 
is
Thus, the average velocity is 2 m/s.
