Physics Calc. 物理計算題 ~20點!!

2015-01-04 10:13 am
求教學~~ Physics Online Homework 急!!!

Q1
http://s30.postimg.org/7gwwvi1rj/Pipes.jpg

Water (viscosity = 8.90 × 10-4 Pa s, density = 1.00 ×103 kg/m3) flows along the pipe shown in the diagram, the diagram is not to scale. The first pipe (at the top) has a radius, R1 = 0.5470 m and the second pipe (at the bottom) has a radius, r2 = 0.2810 m. Water in the first pipe has a pressure of P1 = 1.133 atm and flows at a speed of v1 = 0.3680 ms-1. The height difference, h, between the two pipes is 10 m.

What is the speed of water along the second pipe?__ ms-1

What is the pressure of water in the second pipe?___atm

What is the rate at which the water looses gravitational potential energy?__ W

Q2
http://postimg.org/image/7a1exlnwf/
Two balls collide. The diagram below shows the initial and final velocities of the balls.
In this case:

m1 = 1.88 kg
m2 = 2.15 kg
u1 = 4.91 ms-1
u2 = -2.59 ms-1
v1 = -1.56 ms-1
v2 = ?

What is the initial momentum of the system?__ kgms-1 (Note: if it is to the left enter a negative answer)

What is the final velocity of ball 2?__ ms-1 to the right (enter a negative number if it is to the left)

If the collision lasts for 0.124 s what is the magnitude of the impulse felt by ball 1?__ Ns (right is positive, left is negative)

What average force is experienced by ball 1 during the collision?__ N (right is positive, left is negative)

Select any of the comments below that are correct (there may be more than one correct answer).
Select one or more:
a. Kinetic energy is conserved in this collision so the collision is elastic.
b. The forces experienced by ball 1 and ball 2 have the same magnitude.
c. Momentum is conserved so this is an elastic collision.
d. Ball 1 and ball 2 experience identical forces.
e. Kinetic energy is not conserved in this collision so this is an inelastic collision.
謝謝!!
更新1:

謝謝你的幫忙~~不過我交左first draft 話呢兩條答案不對~~想問問會不會有別的做法? What is the pressure of water in the second pipe?___atm What is the rate at which the water looses gravitational potential energy?__ W

回答 (1)

2015-01-04 10:03 pm
✔ 最佳答案
1. Using continuity equation:
pi.(0.547^2) x 0.368 x 1000 = pi.(0.281^2) x V x 1000
where V is the speed of water along the lower pipe, and pi = 3.14159.....
i.e. V = 1.394 m/s

Using Bernouli's equation:
pressure difference = [(1/2).(1000)(1.394^2 - 0.368^2) + 1000g(10)] Pa
= 100904 Pa = 1.009 atm
where g is the acceleration due to gravity, taken to be 10 m/s^2, and 1 atm = 10^5 Pa
Hence, pressure at lower pipe = (1.133 - 1.009) atm = 0.124 atm

Water flow rate = pi.(0.547^2) x 0.368 x 1000 kg/s = 345.9 kg/s
Hence, rate of loss of potential energy = 345.9g x 10 J/s = 34,590 W

2. Initial momentum = [1.88 x 4.91 + 2.15 x (-2.59)] kg.m/s = 3.66 kg.m/s

1.88 x (-1.56) + 2.15(v2) = 3.66
hence, v2 = 3.066 m/s

Change of momentum of ball 1 = 1.88 x (-1.56 - 4.91) kg.m/s = -12.16 kg.m/s
Hence, impulse = -12.16 N.s

Average force = -12.16/0.124 N = -98.1 N

The correct comments are: (b) and (e)



2015-01-07 16:04:32 補充:
Use Bernouli's equation:
(1/2).(1000).(0.368^2) + (1000)g.(10) + (1.133).(10^5) = (1/2).(1000).(1.394^2) + P2
P2 = [(1/2).(1000).(0.368^2-1.394^2) + (1000g).(10) +1.133x10^5] Pa = 2.32 x 10^5 Pa
i.e. P2 = 2.32 atm


收錄日期: 2021-04-20 15:22:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150104000051KK00008

檢視 Wayback Machine 備份