Higher Physics 2013-2014
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Original post by CSM1996
Hmm idk, try posting a picture of the diagram, it might help the person who is able to help you 
Parallel weight . . .
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Original post by Pennyarcade
Ive had a mind blank and I completely cant remember how to do this one:
Two boys are pulling a car of mass 800 kg
along a level surface with a pair of ropes attached
horizontally as shown below. (both ropes being pulled at 20 degrees from the dotted line in separate directions)
When the pull on each rope is 400 N in the
directions indicated, the acceleration of the car is
0.1ms-1
.
What is the size of the frictional force acting on
the car in the above situation?
A 194N
B 434 N
C 533N
D 672 N
E 832N
thanks.
Two boys are pulling a car of mass 800 kg
along a level surface with a pair of ropes attached
horizontally as shown below. (both ropes being pulled at 20 degrees from the dotted line in separate directions)
When the pull on each rope is 400 N in the
directions indicated, the acceleration of the car is
0.1ms-1
.
What is the size of the frictional force acting on
the car in the above situation?
A 194N
B 434 N
C 533N
D 672 N
E 832N
thanks.
Use the unbalanced force parallel to the horizontal.
I don't know how to put something in a spoiler but this should help if you're not sure.
Fun=ma
F - friction = ma, where F is the force parallel which is "force at an angle x cos(angle to the horizontal)"
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Original post by FizzicsGuy
Use the unbalanced force parallel to the horizontal.
I don't know how to put something in a spoiler but this should help if you're not sure.
Fun=ma
F - friction = ma, where F is the force parallel which is "force at an angle x cos(angle to the horizontal)"
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I don't know how to put something in a spoiler but this should help if you're not sure.
Fun=ma
F - friction = ma, where F is the force parallel which is "force at an angle x cos(angle to the horizontal)"
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Yeah I just did that and I got an answer of 375.89
Original post by Pennyarcade
Yeah I just did that and I got an answer of 375.89
It's late and I've done it quickly, but there is TWO ropes pulling at 400N. Did you account for that?
Original post by TheFOMaster
It's late and I've done it quickly, but there is TWO ropes pulling at 400N. Did you account for that?
Original post by FizzicsGuy
Use the unbalanced force parallel to the horizontal.
I don't know how to put something in a spoiler but this should help if you're not sure.
Fun=ma
F - friction = ma, where F is the force parallel which is "force at an angle x cos(angle to the horizontal)"
Posted from TSR Mobile
I don't know how to put something in a spoiler but this should help if you're not sure.
Fun=ma
F - friction = ma, where F is the force parallel which is "force at an angle x cos(angle to the horizontal)"
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I ended up getting it, my calculator was just being a weirdo
I don't know how to do this question I've attempted it and my working
is below but I got the answer wrong
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(Uploaded in two parts by mistake oops)
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Wooo got 15/16 in my physics test 
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Q2 please 
http://www.sqa.org.uk/pastpapers/papers/papers/2013/H_Physics_all_2013.pdf
How do you do it?
http://www.sqa.org.uk/pastpapers/papers/papers/2013/H_Physics_all_2013.pdf
How do you do it?
Original post by Delenelie
Q2 please
http://www.sqa.org.uk/pastpapers/papers/papers/2013/H_Physics_all_2013.pdf
How do you do it?
http://www.sqa.org.uk/pastpapers/papers/papers/2013/H_Physics_all_2013.pdf
How do you do it?
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Original post by Ewan-Hunter
Use your knowledge to establish the relationship between voltage and light intensity. If the light intensity is proportional to the light output, then with a higher light intensity you'll get a brighter bulb. Would this mean a higher or lower voltage is required?
Spoiler
Thanks very much, got it
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The driver of a car travelling along a level road brakes sharply. The from wheels of the car lock producing the skid marks shows below* as the car comes to rest
(The diagram basically shows that the mark was 32.2m long)
The mass of the car including the driver is 1050kg. It is estimates that the frictional force while skidding is 5210N.
Calculate the speed of the car just before it's wheels looked.
I calculated the acceleration first. which is a=F/m= 4.96ms-1
I then wanted to use one of the equations of motion to work out the time and then use s=d/t to calculate the speed but I don't know what 'u' and 'v' is. What should I do? Thanks
(The diagram basically shows that the mark was 32.2m long)
The mass of the car including the driver is 1050kg. It is estimates that the frictional force while skidding is 5210N.
Calculate the speed of the car just before it's wheels looked.
I calculated the acceleration first. which is a=F/m= 4.96ms-1
I then wanted to use one of the equations of motion to work out the time and then use s=d/t to calculate the speed but I don't know what 'u' and 'v' is. What should I do? Thanks
Original post by Delenelie
The driver of a car travelling along a level road brakes sharply. The from wheels of the car lock producing the skid marks shows below* as the car comes to rest
(The diagram basically shows that the mark was 32.2m long)
The mass of the car including the driver is 1050kg. It is estimates that the frictional force while skidding is 5210N.
Calculate the speed of the car just before it's wheels looked.
I calculated the acceleration first. which is a=F/m= 4.96ms-1
I then wanted to use one of the equations of motion to work out the time and then use s=d/t to calculate the speed but I don't know what 'u' and 'v' is. What should I do? Thanks
(The diagram basically shows that the mark was 32.2m long)
The mass of the car including the driver is 1050kg. It is estimates that the frictional force while skidding is 5210N.
Calculate the speed of the car just before it's wheels looked.
I calculated the acceleration first. which is a=F/m= 4.96ms-1
I then wanted to use one of the equations of motion to work out the time and then use s=d/t to calculate the speed but I don't know what 'u' and 'v' is. What should I do? Thanks
s=d/t assumes constant speed, but the car is decelerating.
In this case, we know he decelerates to rest, so what is the final velocity, v? We know the deceleration (from the frictional force and mass).
We are trying to work out his initial velocity, u.
Original post by Asklepios
s=d/t assumes constant speed, but the car is decelerating.
In this case, we know he decelerates to rest, so what is the final velocity, v? We know the deceleration (from the frictional force and mass).
We are trying to work out his initial velocity, u.
In this case, we know he decelerates to rest, so what is the final velocity, v? We know the deceleration (from the frictional force and mass).
We are trying to work out his initial velocity, u.
I get it now
Thank youOriginal post by Delenelie
Anyone? I assume I need to use the equations of motion?
Yes you do. Just remember to keep the vertical and horizontal values separate.
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