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username1894983
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A car exerts a driving force of 500N when travelling at a constant speed of 72 KMH-1 on a level track. What is the work done in 5 minutes??
According to the marksceme the answer is 2.0x10^5
Can someone explain how to get that that answer. Thanks !
According to the marksceme the answer is 2.0x10^5
Can someone explain how to get that that answer. Thanks !
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Teenie2
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#2
I don't get that answer, what I get is :-
Work done = Force * Distance.
Distance travelled = speed * time = 72 * (5/60) = 6km = 6000m
So, Work done - 500N * 6000m = 3x10^6 Nm
Is that what you've been getting?
Work done = Force * Distance.
Distance travelled = speed * time = 72 * (5/60) = 6km = 6000m
So, Work done - 500N * 6000m = 3x10^6 Nm
Is that what you've been getting?
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Joinedup
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#3
72 km h^-1 = 20 m s^-1
in 300 secs the car travels 6000 m
over that distance the car is exerting constant 500N force
work done = Fs
500 x 6000 = 3x10^6 J
in 300 secs the car travels 6000 m
over that distance the car is exerting constant 500N force
work done = Fs
500 x 6000 = 3x10^6 J
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srata
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Joinedup
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#5
fwiw MS says 3.0 x 10^6 J too
Q23 in paper http://filestore.aqa.org.uk/resource...081-SQP.PDF#24
answer to 23 is A (3.0 x 10^6 J)
in MS http://filestore.aqa.org.uk/resource...081-SMS.PDF#13
Q23 in paper http://filestore.aqa.org.uk/resource...081-SQP.PDF#24
answer to 23 is A (3.0 x 10^6 J)
in MS http://filestore.aqa.org.uk/resource...081-SMS.PDF#13
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Teenie2
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#6
I think you've been reading the wrong bit of the mark scheme Kraixo - the answer is A - 3x10^6J
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username3295148
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#7
The mark scheme states that it is C 2.0*10^6 but when calculated it comes to 3.0*10^6 (A)
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Pangol
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#8
When this specimen paper was first released, there were a couple of errors on the mark scheme, including this one. As others have pointed out, they have since put it right!
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username3295148
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Pangol
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#10
(Original post by MH1234567)
I got this question from physics and maths tutor
I got this question from physics and maths tutor
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Yungkillag
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#11
(Original post by username1894983)
A car exerts a driving force of 500N when travelling at a constant speed of 72 KMH-1 on a level track. What is the work done in 5 minutes??
According to the marksceme the answer is 2.0x10^5
Can someone explain how to get that that answer. Thanks !
A car exerts a driving force of 500N when travelling at a constant speed of 72 KMH-1 on a level track. What is the work done in 5 minutes??
According to the marksceme the answer is 2.0x10^5
Can someone explain how to get that that answer. Thanks !
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Saifur1111
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#12
was also confused by this, was sure my workings were correct, since then the mark scheme has not been updated. From what i see the answer is 3*10^6
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