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Snazarina
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#1
An ice manufacturing machine requires 10 minutes to warm up before the production of ice begins. The mass, in tonnes, of ice produced is directly proportional to the number of hours of production. Given that 20 tonnes of ice are produced when the machine ran for half an hour, find the mass of ice produced when thr machine ran for 7/4 hours.
Ok so i did m=kh (m is mass and h is hours )
20=k 1/2
K=40
When k=40, m= 70 tonnes.
However, the answer says 95 tonnes. Am i missing something here? Can someone help pls.
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Ok so i did m=kh (m is mass and h is hours )
20=k 1/2
K=40
When k=40, m= 70 tonnes.
However, the answer says 95 tonnes. Am i missing something here? Can someone help pls.
Posted from TSR Mobile
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Mr M
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#2
(Original post by Snazarina)
An ice manufacturing machine requires 10 minutes to warm up before the production of ice begins. The mass, in tonnes, of ice produced is directly proportional to the number of hours of production. Given that 20 tonnes of ice are produced when the machine ran for half an hour, find the mass of ice produced when thr machine ran for 7/4 hours.
Ok so i did m=kh (m is mass and h is hours )
20=k 1/2
K=40
When k=40, m= 70 tonnes.
However, the answer says 95 tonnes. Am i missing something here? Can someone help pls.
Posted from TSR Mobile
An ice manufacturing machine requires 10 minutes to warm up before the production of ice begins. The mass, in tonnes, of ice produced is directly proportional to the number of hours of production. Given that 20 tonnes of ice are produced when the machine ran for half an hour, find the mass of ice produced when thr machine ran for 7/4 hours.
Ok so i did m=kh (m is mass and h is hours )
20=k 1/2
K=40
When k=40, m= 70 tonnes.
However, the answer says 95 tonnes. Am i missing something here? Can someone help pls.
Posted from TSR Mobile
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Snazarina
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#3
(Original post by Mr M)
Yes you are failing to take account of the time the machine needs to warm up.
Yes you are failing to take account of the time the machine needs to warm up.
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Mr M
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#4
It is easy to include it the equation though. I'll let you think about it a bit more.
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krisshP
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#5
When 20 tonnes of ice is produced, machines runs for half an hour, so produces for 20 mins which is 1/3 hours.
20=(k)(1/3)
You do the rest now
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Snazarina
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#6
(Original post by Mr M)
You don't need to if you don't want to. Just assume the machine runs for 20 mins instead of 30.
It is easy to include it the equation though. I'll let you think about it a bit more.
You don't need to if you don't want to. Just assume the machine runs for 20 mins instead of 30.
It is easy to include it the equation though. I'll let you think about it a bit more.
Hence m=60 (7/4-1/6)
m=95.
Perfect! Thank you!
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Mr M
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#7
(Original post by Snazarina)
Ok assuming it runs for 20 mins then i've got k=60
Hence m=60 (7/4-1/6)
m=95.
Perfect! Thank you!
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Ok assuming it runs for 20 mins then i've got k=60
Hence m=60 (7/4-1/6)
m=95.
Perfect! Thank you!
Posted from TSR Mobile
If you want a formula that includes the 1/6 it would be:
but I guess you have already realised that.
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Snazarina
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#8
(Original post by Mr M)
Well done.
If you want a formula that includes the 1/6 it would be:
but I guess you have already realised that.
Well done.
If you want a formula that includes the 1/6 it would be:
but I guess you have already realised that.
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raqureshi78
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#9
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#9
I instead do not convert minutes to hours.m = 20, h = running time, so h=20 min because machine requires 10 mins for warm up.m = k.h20= k. 20k= 1now we have running hours 1.75 = 105 min. 10 mins of warm up will be subtracted, we have 95 mins remainingm = k. h.m = 1. 95m = 95 tonnes
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Nawalkhalil
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