Some physics qns!

last SG · 2008-10-29T00:25:50+00:00

Urgent Help!!Calculate the density of the ice at 0 degree...Er..Someone help pls..!! I get 1 g/cm^3 but the answer in the book say 0.909g/cm^3.Since u posted...

skythewood SG

But why u take 11 but not 10?Pretend you did not see the answer how u know is take 11 and not 10 cause in 0 degree they are at 11-10

fact, water expands when it freezes into ice.

hence, it is 11, not 10. 10 is just water at 0 degree

jlzk SG

k i try to digest abit thanks!

skythewood SG

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.

MyNameisNas SG

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).

skythewood SG

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.

i have no idea what you trying to say though. you can find PE by using mgh?

MyNameisNas SG

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)''

skythewood SG

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.

jlzk SG

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

eagle SG

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

jlzk SG

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!!!

wishboy SG

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

davidche SG

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

davidche SG

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?

davidche SG

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

wishboy SG

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

eagle SG

140000N

eagle can help? or also bu zhi suo cuo

which question is that? So many questions I lost liao

wishboy SG

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.

davidche SG

my question 4 in 1st post.

eagle SG

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.)

eagle SG

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.

The question do not need to put the point where GPE = 0

Because we are finding the change in GPE.

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