Erm, sorry.
29.66 is ~70% of 41.9
(29.66/41.9)*100
29.66 is ~70% of 41.9
(29.66/41.9)*100
I tried that...I did my own calculation and got 70.34%, then i entered 70% for my answer and it was wrong. I'm confusedliquidat0r wrote:
Erm, sorry.
29.66 is ~70% of 41.9
(29.66/41.9)*100
wut. oh
Try: (41.9 - 29.66)/41.9 * 100
= 29.21%
Try: (41.9 - 29.66)/41.9 * 100
= 29.21%
huh.....that didn't work either, oh well I just used up all 10 chances to get the question. It's okay though, homework is due in 9 minutes away.liquidat0r wrote:
wut. oh
Try: (41.9 - 29.66)/41.9 * 100
= 29.21%
~Thanks for the help.
No idea then, because:
Percentage difference means to show the difference (subtract one value from another) as a percent of the old value ... so divide by the old value and make it a percentage.
meh.
Percentage difference means to show the difference (subtract one value from another) as a percent of the old value ... so divide by the old value and make it a percentage.
meh.
I need help on some physics again:
1) A satellite is in orbit around a planet. The orbital radius is 32 km and the gravitational acceleration at that height is 3.3 m/s^2. What is the satellite's orbital speed?
~For this problem I got 10.276186 m/s but I don't know if that is right or not.
2) A 1100-kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m. What is the average braking force acting on the car?
~This one looks like it is easy but I keep getting really small numbers and this problem has a multiple choice answer: (a) -690 N (b) -340 N (c) -410 N (d) -550 N
3) A block is on a frictionless table, on Earth. The block accelerates at 3.6 m/s^2 when a 90 N horizontal force is applied to it. The block and table are set up on the Moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s^2. A horizontal force of 45 N is applied to the block when it is on the Moon. The acceleration imparted to the block is closest to: (a) 1.6 m/s^2 (b) 2.2 m/s^2 (c) 2.3 m/s^2 (d) 2.0 m/s^2 (e) 1.8 m/s^2
~This one I worked out and came up with 1.6 m/s^2 but forgot how I did it and now I'm not sure if this answer is right
1) A satellite is in orbit around a planet. The orbital radius is 32 km and the gravitational acceleration at that height is 3.3 m/s^2. What is the satellite's orbital speed?
~For this problem I got 10.276186 m/s but I don't know if that is right or not.
2) A 1100-kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m. What is the average braking force acting on the car?
~This one looks like it is easy but I keep getting really small numbers and this problem has a multiple choice answer: (a) -690 N (b) -340 N (c) -410 N (d) -550 N
3) A block is on a frictionless table, on Earth. The block accelerates at 3.6 m/s^2 when a 90 N horizontal force is applied to it. The block and table are set up on the Moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s^2. A horizontal force of 45 N is applied to the block when it is on the Moon. The acceleration imparted to the block is closest to: (a) 1.6 m/s^2 (b) 2.2 m/s^2 (c) 2.3 m/s^2 (d) 2.0 m/s^2 (e) 1.8 m/s^2
~This one I worked out and came up with 1.6 m/s^2 but forgot how I did it and now I'm not sure if this answer is right
I think the correct answer is: (e) 1.8m/s^2TheunforgivenII wrote:
3) A block is on a frictionless table, on Earth. The block accelerates at 3.6 m/s^2 when a 90 N horizontal force is applied to it. The block and table are set up on the Moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s^2. A horizontal force of 45 N is applied to the block when it is on the Moon. The acceleration imparted to the block is closest to: (a) 1.6 m/s^2 (b) 2.2 m/s^2 (c) 2.3 m/s^2 (d) 2.0 m/s^2 (e) 1.8 m/s^2
~This one I worked out and came up with 1.6 m/s^2 but forgot how I did it and now I'm not sure if this answer is right
The table is frictionless - therefore the difference in gravitational force, experienced on the earth and on the moon, is irrelevant - all you need to take into account is the force applied to the block - 45N is 90N/2, so the acceleration must be 3.6m/s^2/2, i.e. 1.8m/s^2 - this can be confirmed by calculating the mass of the block (from the 'on the earth' data) and then working out the acceleration...
For circular motion, a = v2/r, where:TheunforgivenII wrote:
I need help on some physics again:
1) A satellite is in orbit around a planet. The orbital radius is 32 km and the gravitational acceleration at that height is 3.3 m/s^2. What is the satellite's orbital speed?
~For this problem I got 10.276186 m/s but I don't know if that is right or not.
a = acceleration
v = tangential velocity
r = radius of circle/orbit
Rearranging gives v = √(a x r),
BEFORE YOU PUT THE VALUES INTO THE EQUATION, ENSURE THAT ALL UNITS ARE SI UNITS. This is why you are wrong. You didn't change Km to m.
Thus,
v = √(32,000 x 3.3)
v = √(105600)
v = 324.9615 m/s
In my physics course last year at uni, we would get our arses kicked for leaving it like that. Both values you revieced were to 2 significant figures, so technically your answer should be to 2 significant figures -> 320m/s.
Okay I think I understand now, thanks for the help.Scorpion0x17 wrote:
I think the correct answer is: (e) 1.8m/s^2TheunforgivenII wrote:
3) A block is on a frictionless table, on Earth. The block accelerates at 3.6 m/s^2 when a 90 N horizontal force is applied to it. The block and table are set up on the Moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s^2. A horizontal force of 45 N is applied to the block when it is on the Moon. The acceleration imparted to the block is closest to: (a) 1.6 m/s^2 (b) 2.2 m/s^2 (c) 2.3 m/s^2 (d) 2.0 m/s^2 (e) 1.8 m/s^2
~This one I worked out and came up with 1.6 m/s^2 but forgot how I did it and now I'm not sure if this answer is right
The table is frictionless - therefore the difference in gravitational force, experienced on the earth and on the moon, is irrelevant - all you need to take into account is the force applied to the block - 45N is 90N/2, so the acceleration must be 3.6m/s^2/2, i.e. 1.8m/s^2 - this can be confirmed by calculating the mass of the block (from the 'on the earth' data) and then working out the acceleration...
Damn Adam, you suck. I didn't even need to go through the rest of the pages.
I suck?
Oh, I see what you're talking about now. Yes, I failed at working out a percentage difference ... or whatever it was, can't remember now.
Oh, I see what you're talking about now. Yes, I failed at working out a percentage difference ... or whatever it was, can't remember now.
Last edited by liquidat0r (2009-01-26 15:24:44)
Oh i see what I did wrong, I used the same equation as you did but I didn't use the number 32,000, I used v=squareroot(32*3.3)mcminty wrote:
For circular motion, a = v2/r, where:TheunforgivenII wrote:
I need help on some physics again:
1) A satellite is in orbit around a planet. The orbital radius is 32 km and the gravitational acceleration at that height is 3.3 m/s^2. What is the satellite's orbital speed?
~For this problem I got 10.276186 m/s but I don't know if that is right or not.
a = acceleration
v = tangential velocity
r = radius of circle/orbit
Rearranging gives v = √(a x r),
BEFORE YOU PUT THE VALUES INTO THE EQUATION, ENSURE THAT ALL UNITS ARE SI UNITS. This is why you are wrong. You didn't change Km to m.
Thus,
v = √(32,000 x 3.3)
v = √(105600)
v = 324.9615 m/s
In my physics course last year at uni, we would get our arses kicked for leaving it like that. Both values you revieced were to 2 significant figures, so technically your answer should be to 2 significant figures -> 320m/s.
This.TheunforgivenII wrote:
2) A 1100-kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m. What is the average braking force acting on the car?
~This one looks like it is easy but I keep getting really small numbers and this problem has a multiple choice answer: (a) -690 N (b) -340 N (c) -410 N (d) -550 N
I'll type out the working. Watch this space.
btw, lol Zimmer
- - - - - - - -
1. Determine the acceleration (deceleration).
v2 = u2 + 2as, where
v = final velocity
u = initial velocity
a = acceleration
s = displacement
You are given that v = 0 (the car stops), u = 27m/s and s=578.0 Thus,
02 = 272 + 2 x 578 x a
0 = 729 + 1156a
1156a = -729
a= -729/1156 = -0.6306 m/s2
2. Using acceleration, determine the stopping force.
F = ma
Given, m = 1100, a = -0.6306
F = 1100 x -0.6303
F = -693.66N ~ -690N
Just use Newton's second law in the form:2) A 1100-kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m. What is the average braking force acting on the car?
F = ma
And this equation of motion:
v2 = u2 + 2as
Where:
m = mass
F = braking force
v = final velocity
u = initial velocity
s = displacement
You end up with an equation like this:
F = (mv2)/2s
The answer is ~690 N.
---
By the time you read this, minty. You will have wasted quite a few minutes typing out working
Last edited by liquidat0r (2009-01-26 15:25:44)
wow....I feel stupid, I was trying to figure out what the acceleration and wasted a whole time trying to solve for time because at first I wanted to just plug everything in F = ma. Never once thought about using the equation for motionliquidat0r wrote:
Just use Newton's second law in the form:2) A 1100-kg car traveling at 27 m/s starts to decelerate and comes to a complete stop in 578.0 m. What is the average braking force acting on the car?
F = ma
And this equation of motion:
v2 = u2 + 2as
Where:
m = mass
F = braking force
v = final velocity
u = initial velocity
s = displacement
You end up with an equation like this:
F = (mv2)/2s
The answer is ~690 N.
Last edited by TheunforgivenII (2009-01-26 15:30:10)
I guess you could have done it using this, too:
Work done = Fs
Work done = 0.5mv2
And then F = (mv2)/2s
But that's less obvious and would require more written explanation (conservation of energy/momentum). I only noticed it because the equation we ended up has the equation for KE in it.
Work done = Fs
Work done = 0.5mv2
And then F = (mv2)/2s
But that's less obvious and would require more written explanation (conservation of energy/momentum). I only noticed it because the equation we ended up has the equation for KE in it.
liquidat0r wrote:
By the time you read this, minty. You will have wasted quite a few minutes typing out working
But it's a motion question to begin with..TheunforgivenII wrote:
wow....I feel stupid, I was trying to figure out what the acceleration and wasted a whole time trying to solve for time because at first I wanted to just plug everything in F = ma. Never once thought about using the equation for motion
yeah....that's another careless mistake I made which was not reading the question more carefully...mcminty wrote:
But it's a motion question to begin with..TheunforgivenII wrote:
wow....I feel stupid, I was trying to figure out what the acceleration and wasted a whole time trying to solve for time because at first I wanted to just plug everything in F = ma. Never once thought about using the equation for motion
Always read the question at least two times before answering.TheunforgivenII wrote:
yeah....that's another careless mistake I made which was not reading the question more carefully...mcminty wrote:
But it's a motion question to begin with..TheunforgivenII wrote:
wow....I feel stupid, I was trying to figure out what the acceleration and wasted a whole time trying to solve for time because at first I wanted to just plug everything in F = ma. Never once thought about using the equation for motion
Yup, and determine what type of questions it is (what they want to know), as this will govern the equations you use. Also, write down the given data, and convert it to SI units.Scorpion0x17 wrote:
Always read the question at least two times before answering.TheunforgivenII wrote:
yeah....that's another careless mistake I made which was not reading the question more carefully...mcminty wrote:
But it's a motion question to begin with..
When you come up with an answer, think "does this really make sense?". I initially made the same mistake as you for that orbital velocity question, but when I came up with ~10m/s I thought that it didn't look right, and had to be wrong.. so I went back and checked all the units, etc (thus finding your error).
for me I will most certainly have to read every question more than 4 timesScorpion0x17 wrote:
Always read the question at least two times before answering.TheunforgivenII wrote:
yeah....that's another careless mistake I made which was not reading the question more carefully...mcminty wrote:
But it's a motion question to begin with..
Last edited by TheunforgivenII (2009-01-26 15:50:55)
1) A 1.3-kg object is moving in the x direction at 17.4 m/s. Just 3.7 s later, it is moving at 28.5 m/s at 32.7° to the x axis. What are the magnitude and direction of the force applied during this time?
I split this problem into two sections of x and y:
x-components: y-components:
initial velocity = 17.4 m/s initial velocity = 0
final velocity = ? final velocity = ?
time = 3.7 seconds time = 3.7 seconds
acceleration = ? acceleration = ?
so what I did to find the final velocities was 28.5(cos32.7)) for the x-component and 28.5(sin32.7)) for the y-components. Then I found acceleration buy dividing the change in velocity over time on both x and y components. Next I plugged everything into F = ma for x and y. After that I used the pythagorean theorem to find the total force, F2 = Fx2 + Fy2, and got 12.50 N but this answer is wrong, I don't know where I messed up. Also I can't find the answer for θF
2) You wish to row straight across a 50-m-wide river. You can row at a steady 1.6 m/s relative to the water, and the river flows at 0.5 m/s.
for this one I tried to use vrow(cosx)-vriver = 0. This advice was given to me by my TA and I still can't figure it out
I split this problem into two sections of x and y:
x-components: y-components:
initial velocity = 17.4 m/s initial velocity = 0
final velocity = ? final velocity = ?
time = 3.7 seconds time = 3.7 seconds
acceleration = ? acceleration = ?
so what I did to find the final velocities was 28.5(cos32.7)) for the x-component and 28.5(sin32.7)) for the y-components. Then I found acceleration buy dividing the change in velocity over time on both x and y components. Next I plugged everything into F = ma for x and y. After that I used the pythagorean theorem to find the total force, F2 = Fx2 + Fy2, and got 12.50 N but this answer is wrong, I don't know where I messed up. Also I can't find the answer for θF
2) You wish to row straight across a 50-m-wide river. You can row at a steady 1.6 m/s relative to the water, and the river flows at 0.5 m/s.
for this one I tried to use vrow(cosx)-vriver = 0. This advice was given to me by my TA and I still can't figure it out
Question 2
Hopefully you can read/decipher my writing/working.
Would type it out and make it look nicer and easier to follow but I don't have time.
The v's with subscripts are meant to be like these:
v1y = Initial velocity in y direction.
v1x = Initial velocity in x direction.
v2y = Final velocity in y direction.
v2x = Final velocity in x direction.
(Same goes for the Forces (F) and accelerations (a).
Hopefully you can read/decipher my writing/working.
Would type it out and make it look nicer and easier to follow but I don't have time.
The v's with subscripts are meant to be like these:
v1y = Initial velocity in y direction.
v1x = Initial velocity in x direction.
v2y = Final velocity in y direction.
v2x = Final velocity in x direction.
(Same goes for the Forces (F) and accelerations (a).
adams got nice writing.
Small hourglass island
Always raining and foggy
Use an umbrella
Always raining and foggy
Use an umbrella