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antoniya [11.8K]
2 years ago
11

Can and object have a negative position and a positive velocity? Or vice versa, a positive position and a negative velocity? Exp

lain your reasoning, do not just give a simple sentence answer.​
Physics
1 answer:
zavuch27 [327]2 years ago
4 0
Imagine a ball is moving on the following horizontal line.

. . . . . . . . . . . . . . . . . . . O. . . . . . . . . . . . . . . . . .

Take right as positive. O is the starting point of the ball. Denote the ball by o.

. . . . . . . . . . . . . . . . . . . O. . . . . . . ... . . o . . . . . .

Assume the ball is moving to the right. It has positive displacement since it is on the right of O, and positive velocity since its positive displacement is increasing.

.ñ

. . . . . . . . . . . . . . . . . . . O. . . . o . . . . . . . . . . . . .

Now the ball is returning to O. It still has positive displacement since its current position is still on the right of O. However, its velocity is negative since its positive displacement is decreasing and the direction of the velocity vector points left, which is the negative side.

By now you should be able to come up with a scenario where the ball has negative displacement and positive velocity.

You can observe the same phenomenon in daily life. Say, as a stretched spring bounces to its starting position, if we let the returning direction be positive, the string has negative displacement since it is on the negative direction, but has positive velocity. Bungee jump can also used to illustrate the phenomenon.
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A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1
stealth61 [152]

Answer:

m2=3.2722lbm/s

Explanation:

Hello!

To solve this problem follow the steps below

1. Find water densities and entlapies  in all states using thermodynamic tables.

note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)

through prior knowledge of two other properties, such as pressure and temperature.

D1=Density(Water;T=50;x=0) =62.41 lbm/ft^3

D2=Density(Water;T=120;x=0) =61.71 lbm/ft^3

D3=Density(Water;T=80;x=0) =62.21 lbm/ft^3

h1=Enthalpy(Water;T=50;x=0) =18.05 BTU/lbm

h2=Enthalpy(Water;T=120;x=0) =88  BTU/lbm

h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm

2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit

m1+m2=m3

3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out

m1h1+m2h2=m3h3

18.05(m1)+88(m2)=48.03(m3)

divide both sides of the equation by 48.03

0.376(m1)+1.832(m2)=m3

4. Subtract the equations obtained in steps 3 and 4

m1            +      m2       =  m3

-

0.376m1   +  1.832(m2) =m3

--------------------------------------------

0.624m1-0.832m2=0

solving for m2

(0.624/0.832)m1=m2

0.75m1=m2

5. Mass flow is the product of density by velocity across the cross-sectional area

m1=(D1)(A)(v1)

internal Diameter for  2" Sch 40=2.067in=0.17225ft

A=\frac{\pi }{4} D^2=\frac{\pi }{4} (0.17225)^2=0.0233ft^2

m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s

6. use the equation from step 4 to find the mass flow in 2

0.75m1=m2

0.75(4.3629)=m2

m2=3.2722lbm/s

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Explanation:

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