物理讨论专区

cheesit92

work done= gain in kinetic energy,
(&#189I2w2^2 - (&#189I1w1^2  =(&#189mr2^2 w2^2  - (&#189mr1^2 w1^2
                                      =(&#189(0.3)(0.3)^2 (27.78)^2 -(&#189(0.3)(0.5)^2 (10)^2
                                      =6.6668J

vick5821

Gravitational Ques!!
1.A rocket is launched vertically from the surface of the Earth at speed 25 km s-1. Determine its speed when it escapes from the gravitational field of the Earth.

How ar??
URGENT!!!!

vick5821

A particle moving along x axis in SHM starts from the origin at t=0 and moves to the right. The amplitude of the its motion is 2.00cm and the frequency is 1.50 Hz.
Calculate the maximum acceleration and the earliest time (t>0) at which the particle has this acceleration

Winterblade

a=-(w^2)(x)
Amplitude peaked at 0.02m
Frequency= 1.50Hz
a=-(2piF)(0.02)
  =-1.78ms^-2

T=period for one complete cycle.
  =2pi/w
  =2pi/2piF
  =1/1.5
  =2/3s

Earliest peak occur at 1/4 T
1/4x2/3= 1/6s

I am not certain about that...Sorry if i make any mistakes

vick5821

yes..I think u r correct..the answer given is wrong..thx for helping!!

Winterblade

>.<...What is the answers?

nkrealman

关于 static...
ladder with centre of mass resting on wall，斜放着。
为什么頭尾的normal reaction 是斜上去？

Winterblade

回复 430# nkrealman


    I think is due to the rough surface of the floor and wall....

whyyie

回复 430# nkrealman

斜上去? 不是perpendicular吗?

Winterblade

I thought perpendicular normal reaction is for smooth surface?

Log

关于 static...
ladder with centre of mass resting on wall，斜放着。
为什么頭尾的normal reaction 是 ...
nkrealman 发表于 31-8-2010 03:14 PM

可以这样想for ladder resting on a rough surface
actually the reaction firstly is acting perpendicularlly, but there exists a frictional force,so by resolving, u will get an inclined reaction force
for smooth surface,the  reaction is normal coz no frictional force.

Winterblade

回复 434# Log

vick5821

the answer for the ques I postred is 1/2s, i think its wrong la..haha
hey, another question here

two waves in a long string are given by
y1 = 0.02m sin ( 40t - x/2) and y2 = 0.02m sin (40t + x/2)
where y1,y2 and x are in metres and t is in seconds
Calculate the maximum displacement of an element in the string at x=0.40m

I dun quite understand this ques..how to do??thanks for the help!!

Log

the answer for the ques I postred is 1/2s, i think its wrong la..haha
hey, another question here

...
vick5821 发表于 2-9-2010 12:51 PM

y1=0.02sin(40t-x/2)
y2=0.02sin(40t+x/2)
resultant wave in the string y=y1+y2
                                         =0.02sin(40t-x/2) + 0.02sin(40t+x/2)
                                         = 0.02 (2  sin 40t cos x/2)
                                        y = 0.04  cos(x/2) sin (40t)
    amplitude=0.04 cos x/2
x=0.40m , so, max diplacement= 0.04 cos (0.40/2)= 0.04 m

vick5821

I dun understand why do like that?? why just take the front part of the equation??

Log

就好像y=a sin bt ,    a=amplitude ,  (y and t are variables)
  etc.      y=2a/c sin (bt +d)               2a/c = amplitude
              y= 2 cos x sin bt             ,  2 cos x=amplitude
so, y= 0.04 cos (x/2 ) sin 40t
0.04 cos (x/2) = amplitude ,   x=distance of a particle measured from the origin = constant

Winterblade

回复 439# Log


    Uhmm how about  resultant wave eqn is y= cos 2(pi)t ((f1-f2)/2)  sin 2(pi)t ((f1+f2)/2)....What is the amplitude?

Log

回复 440# Winterblade
amplitude= (  f1&#178; - f2&#178; )/8
simplify, u obtain  y=  ( f1&#178; - f2&#178; )/8 . sin  4(pi)t

vick5821

so the stationary wave equation is actually for the particlar wave de?? and if we wan to calculate for the displacement of particular element at particular distance, we sub nto cos kx?? how bout the time??

vick5821

2cos x = amplitude then y is what??