Originally posted by skythewood:
i have no idea what you trying to say though. you can find PE by using mgh?
Not in this case as if there is a question that needs to apply Pe=mgh. at the end of the question it would state smtg like ''(gravity= 10N/Kg)''
Originally posted by MyNameisNas:
Not in this case as if there is a question that needs to apply Pe=mgh. at the end of the question it would state smtg like ''(gravity= 10N/Kg)''
oh my god your reason is so good that i bow down to your brilliance.
PHYSICS HELP~!
Explain what is it meant by the frequency is 27MHZ.
Dunno why TYS answer says 27X 10^6 of the wave occur per second.Some one explain pls x.x
M stands for mega, stands for 10^6
27MHz = 27 * 10^6 Hz
and 1 Hz stands for 1 cycle/wave per second
Note: frequency is the number of cycles of the waves in one second
Originally posted by eagle:M stands for mega, stands for 10^6
27MHz = 27 * 10^6 Hz
and 1 Hz stands for 1 cycle/wave per second
Note: frequency is the number of cycles of the waves in one second
Thanks!!!
Originally posted by skythewood:ok, now back to the other q... what is the final answer according to the answer sheet? i have no idea where the t = 2s came from.
Since the dist. travelled before coming to rest is 40m,
Let t be the time taken before it comes to rest.
Area under graph = 40m
½ * 40 * t = 40m (Initial speed is 40m/s)
t = 2s
Originally posted by skythewood:sorry, was talking on the phone with girlfriend.
just give me the final answer.
t=2s? where did that come from?
140000N
eagle can help? or also bu zhi suo cuo
Originally posted by skythewood:oh my god your reason is so good that i bow down to your brilliance.
LOL. so the pt is must put gravity= 10N/kg?
Originally posted by wishboy:Since the dist. travelled before coming to rest is 40m,
Let t be the time taken before it comes to rest.
Area under graph = 40m
½ * 40 * t = 40m (Initial speed is 40m/s)
t = 2s
I still think there is some error. Using this method of getting t=2s, you can get the deceleration of 40-0/ 2 = 20 and F=MA and you DO get the answer.
However, using the energy way, to get 140000 means that one must equate kinetic energy with workdone, ignoring the energy loss due to the slope.
560000 = force * 40m
force = 140000N
Originally posted by davidche:I still think there is some error. Using this method of getting t=2s, you can get the deceleration of 40-0/ 2 = 20 and F=MA and you DO get the answer.
However, using the energy way, to get 140000 means that one must equate kinetic energy with workdone, ignoring the energy loss due to the slope.
560000 = force * 40m
force = 140000N
560000/40 = 14000, not 140000 lol
Originally posted by davidche:140000N
eagle can help? or also bu zhi suo cuo
which question is that? So many questions I lost liao
Originally posted by eagle:which question is that? So many questions I lost liao
A car of mass 700kg with initial kinetic energy of 560KJ and initial speed of 40m/s approaches a slope of height 3m. Note that the car does not brake.
a) Find the change in gravitational potential energy.(this is not impt i am just posting a ''fuller'' qn.)
b) Find the average frictional force on the car if the car comes to rest after 40m.
my question 4 in 1st post.
Originally posted by wishboy:A car of mass 700kg with initial kinetic energy of 560KJ and initial speed of 40m/s approaches a slope of height 3m. Note that the car does not brake.
a) Find the change in gravitational potential energy.(this is not impt i am just posting a ''fuller'' qn.)
b) Find the average frictional force on the car if the car comes to rest after 40m.
a) Change in gpe = mgh = 700 * 10 * 3 = 21 kJ
b) This is assuming the slope is 3 m in height, and 4m in distance (slope). No graphic or anything around...
Using conservation of energy,
WD against friction = initial kinetic energy - increase in gpe
F * 40 = 560 kJ - 21kJ
F * 40 = 539000
F = 14000 N (2 s.f.)
Originally posted by MyNameisNas:A car of mass 700kg with initial kinetic energy of 560KJ and initial speed of 40m/s approaches a slope of height 3m. Note that the car does not brake.
a) Find the change in gravitational potential energy-(Pe).(this is not impt i am just posting a ''fuller'' qn.)
Is that how the actual question is? the confusing part is the question did not put the point of Pe. Basically at 3m Pe=0j and at 0m Pe is 560KJ.
From wiki-cos i lazy to type- the law of conservation of energy states that energy can neither be created nor destroyed, it can only be changed from one form to another or transferred from one body to another, but the total amount of energy remains constant (the same).
The question do not need to put the point where GPE = 0
Because we are finding the change in GPE.
phail. XD
Thats what i did. only way to get the ans is by ignoring the decrease in energy due to GPE change
upslope, 13500N
downslope, 14500N
cannot be 140 000N. that is overkill. i vote for your initial working is correct.
Originally posted by eagle:a) Change in gpe = mgh = 700 * 10 * 3 = 21 kJ
b) This is assuming the slope is 3 m in height, and 4m in distance (slope). No graphic or anything around...
Using conservation of energy,
WD against friction = initial kinetic energy - increase in gpe
F * 40 = 560 kJ - 21kJ
F * 40 = 539000
F = 14000 N (2 s.f.)
err... if 2.s. f. i think is 13000N.
anyway, i vote for answer is wrong.
probably answer used 2 s.f. all the way in their calculations
There's zero graphics. We do not know whether all of the 3 m is indeed travelled
ok see the answer cannot be wrong. Because if you use
Area under graph = 40m
½ * 40 * t = 40m (Initial speed is
40m/s)
t = 2s
F = 700 * (40-0)/2
= 14000
got graph meh? Dun see any graphics leh...
Originally posted by davidche:ok see the answer cannot be wrong. Because if you use
Area under graph = 40m
½ * 40 * t = 40m (Initial speed is 40m/s)
t = 2s
F = 700 * (40-0)/2
= 14000
14000 or 140000?
Originally posted by eagle:got graph meh? Dun see any graphics leh...
he meant the velocity time graph he imagined himself.
sorry it is 14kN all the way.
oh. anyway, i think you are right, and the question need to take into account the change in GPE.
no other thoughts.