The conservation of the mass of fluid through two sections (be they A1 and A2) of a conduit (pipe) or current tube establishes that the mass that enters is equal to the mass that exits. Mathematically the input flow must be the same as the output flow,
The definition of flow is given by
Where
V = Velocity
A = Area
The units of the flow of flow are cubic meters per second, that is to say that if there is a continuity, the volume of input must be the same as that of output, what changes if the sections are modified are the proportions of speed.
In this way
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
Answer:
C. The block and the sphere would have the same weight.
Explanation:
If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.
Answer:
puck decelerates due to the kinetic frictional force μk mg
Explanation:
given data
total distance = 12 m
coefficient of kinetic friction = 0.28
solution
we will apply equation of motion that is
v² - u² = 2 × a × s ................1
we know acceleration will be
a = 
Then we have
Force = mass × acceleration .................2
m × = -μk mg
The puck decelerates due to the kinetic frictional force μk mg
and frictional force is negative as it opposes the motion.
so we get initial velocity of the puck which is strike.
