Foo - Please Help me with Physics homework!!!

It's due tonight at 11:11 PM EST. Please help me

http://www.webassign.net/v4cgiecwierzbinski@southforsyth.ga/student.pl?r=20080923212014ecwierzbinski@southforsyth.ga682189109

Click on Linear Motion - Speed Velocity link on left

lol, webassign.

I had two semesters of doing physics homework on there. I'm glad I'm done with it. :p

btw, no one will be able to log in and see the assignment unless they are enrolled in your class. Maybe you could just tell us the problem?

http://www.ambrosiasw.com/~andrew/funny/noob.jpg

I'm sure it will help when they check the web logs and see a bunch of hits linked from bikeforums.net. And considering that your user name appears to be embedded in the link, they'll know it was you too. IBepicFail.

It's due tonight at 11:11 PM EST. Please help me

http://www.webassign.net/v4cgiecwierzbinski@southforsyth.ga/student.pl?r=20080923212014ecwierzbinski@southforsyth.ga682189109

Click on Linear Motion - Speed Velocity link on left


If you jump from a high window, you won't have to turn in the homework and we should be able to measure the speed and velocity of your descent.

Two suggestions:
1. Buy textbook.
2. Do your own homework.

I'll help you with physics, but only if you don't make me go out of my way to find the problem and you show your work and explain to me why you're stuck.

Here's a couple hints:
F=ma
v=v0 + at

Here's another hint: Draw a good diagram and label it.

I'll help you with physics, but only if you don't make me go out of my way to find the problem and you show your work and explain to me why you're stuck.

+1. I like physics, I pretty much rock at it. Unless you're above first/second level college physics and into theoretical, in which case I know enough to sound smart. But I doubt you would decide to post your postulations on string theory in Foo.

physics II 1988, I 1989...calc based 1989....ugh.

these are my problems

1. [GPhys05 2.037.MC.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
The following quantities describe location or its change: position, distance, and displacement. Which are vectors?
displacement
none of these
position
distance



2. [GPhys05 2.049.MP.] --/1 points Saved Work | Show Details Notes
part score total submissions
1 -- 1 0/3
A bike travels at a constant speed of 3.2 m/s for 8 s. How far does it go?
m

3. [GPhys05 3.020.PP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
If a car accelerates from rest at a constant 4.4 m/s2, how long will it take for the car to reach a velocity of 26 m/s?
s

4. [GPhys05 3.021.PP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
A car slows from 17 m/s to 3.6 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is traveling at 3.6 m/s?
s

5. [GPhys05 3.023.PP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
Use dimensional analysis to convert an airplane's speed of (75 m/s) to km/h.
km/h

6. [GPhys05 3.027.PP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
A race car travels on a racetrack at 36 m/s and slows at a constant rate to a velocity of 17 m/s over 11 s. How far does it move during this time?
m

7. [GPhys05 3.029.PP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
For this problem, sketch the situation, assign variables, create a motion diagram, and then develop a mathematical model. (Do this on paper. Your instructor may ask you to turn in this work.)

A bike rider pedals with constant acceleration to reach a velocity of 7.6 m/s over a time of 4.5 s. During the period of acceleration, the bike's displacement is 19 m. What was the initial velocity of the bike?
m/s

8. [GPhys05 3.035.SR.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
If you were given initial and final velocities and the constant acceleration of an object, and you were asked to find the displacement, what equation would you use?
a = (vf - vi)/t
v = a multiplied by t
vf2 = vi2 + 2ad
d = d0 + 1/2at


9. [GPhys05 3.039.SR.] --/2 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
-- -- 2 --
An airplane starts from rest and accelerates at a constant 2.60 m/s2 for 34.0 s before leaving the ground.
(a) How far did it move?
m

(b) How fast was the airplane going when it took off?
m/s


10. [GPhys05 3.042.PP.] --/2 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
-- -- 2 --
A brick is dropped from a high scaffold.

(a) What is its velocity after 3.3 s?
m/s

(b) How far does the brick fall during this time?
m



11. [GPhys05 3.045.PP.] --/2 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
-- -- 2 --
A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is caught at the same distance above the ground.

(a) How high does the ball rise?
m

(b) How long does the ball remain in the air? (Hint: The time it takes the ball to rise equals the time it takes to fall.)
s



12. [GPhys05 3.061.MC.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
What quantity is represented by the area under a velocity-time graph?
The change in time.
The change in displacement.
The change in acceleration.
The change in velocity.


13. [GPhys05 3.072.AC.] --/2 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
-- -- 2 --
Use the graph in Figure 5-20 and determine the time interval during which the acceleration is largest and the time interval during which the acceleration is smallest.


Figure 5-20

Largest acceleration
t = 0 to t = 0.5 s
t = 5.2 s to t = 6.0 s
t = 31 s to t = 32 s
t = 10.5 s to t = 11.0 s


Smallest acceleration
t = 0 to t = 0.5 s
t = 31 s to t = 32 s
t = 10.5 s to t = 11.0 s
t = 5.2 s to t = 6.0 s


14. [GPhys05 3.076.AC.] --/2 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
-- -- 2 --
The value of g on the moon is 1/6 of its value on Earth.

(a) Would a ball that is dropped by an astronaut hit the surface of the moon with a smaller, equal, or larger speed than that of a ball dropped from the same height to Earth?
The ball will hit the moon with a smaller speed because the acceleration due to gravity is more on the moon.
The ball will hit the moon with a smaller speed because the acceleration due to gravity is more on the moon.
The ball will hit the moon with a smaller speed because the acceleration due to gravity is less on the moon.
The ball will hit the moon with a larger speed because the acceleration due to gravity is less on the moon.


(b) Would it take the ball more, less, or equal time to fall?
The ball will take less time.
The ball will take more time.
The ball will take equal time.



15. [GPhys05 3.078.AC.] --/3 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
3 -- 1 0/3
-- -- 3 --
Rock A is dropped from a cliff and rock B is thrown upward from the same position.

(a) When they reach the ground at the bottom of the cliff, which rock has a greater velocity?
Rock A hits the ground with a greater velocity.
Rock B hits the ground with a greater velocity.

(b) Which has a greater acceleration?
Rock B has a greater acceleration.
Rock A has a greater acceleration.
They have the same acceleration.

(c) Which arrives first?
Rock A reaches the ground first.
Both rocks reach the ground at the same time.
Rock B reaches the ground first.



16. [GPhys05 3.082.MP.] --/4 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
3 -- 1 0/3
4 -- 1 0/3
-- -- 4 --
Refer to Figure 3-19 to find the acceleration of the moving object at each of the following times.

Figure 3-19

(a) during the first 5.0 s of travel
m/s2

(b) between 5.0 s and 10.0 s
m/s2

(c) between 10.0 s and 15.0 s
m/s2

(d) between 20.0 s and 25.0 s
m/s2



17. [GPhys05 3.083.MP.] --/7 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
3 -- 1 0/3
4 -- 1 0/3
5 -- 1 0/3
6 -- 1 0/3
7 -- 1 0/3
-- -- 7 --
Plot a velocity-time graph using the information in Table 3-4, and answer the following questions.

Table 3-4

(a) During what time interval is the object speeding up?
s to s

During what time interval is the object Slowing down?
s to s

(b) At what time does the object reverse direction?
s

(c) How does the average acceleration of the object in the interval between 0 s and 2 s differ from the average acceleration in the interval between 7 s and 12 s?
m/s2 between 0 and 2 s.


18. [GPhys05 3.086.MP.] --/2 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
-- -- 2 --
A supersonic jet flying at 139 m/s experiences uniform acceleration at the rate of 23.1 m/s2 for 18.0 s.

(a) What is its final velocity?
m/s

(b) The speed of sound in air is 331 m/s. What is the plane's speed in terms of the speed of sound?
times the speed of sound



19. [GPhys05 3.087.MP.] --/4 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
3 -- 1 0/3
4 -- 1 0/3
-- -- 4 --
Refer to Figure 5-26 to find the distance traveled during the following time intervals.

Figure 5-26

(a) t = 0 s and t = 5 s
m

(b) t = 5 s and t = 10 s
m

(c) t = 10 s and t = 15 s
m

(d) t = 0 s and t = 25 s
m



20. [GPhys05 3.090.MP.] --/2 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
2 -- 1 0/3
-- -- 2 --
A race car can be slowed with a constant acceleration of -11 m/s2.
(a) If the car is going 48 m/s, how many meters will it travel before it stops?
m

(b) How many meters will it take to stop a car going half as fast?
m


21. [GPhys05 3.091.MP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
A car is traveling 26 m/s when the driver sees a child standing on the road. He takes 0.8 s to react, then steps on the brakes and slows at 8.0 m/s2. How far does the car go before it stops?
m

22. [GPhys05 3.097.MP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
Suppose an astronaut drops a feather from 1.8 m above the surface of the Moon. If the acceleration due to gravity on the Moon is 1.62 m/s2 downward, how long does it take the feather to hit the Moon's surface?
s

23. [GPhys05 3.101.MP.] --/1 points No Response | Show Details Notes
part score total submissions
1 -- 1 0/3
You throw a ball upward from a window at a speed of 1.7 m/s. How fast will it be moving when it hits the sidewalk 3.0 m below?
m/s

Never mind... you seem more interested in having someone do your homework for you than in learning physics. Good luck.

No please. Ok I did about 10 of them got them right. But im stuck on this one

A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is caught at the same distance above the ground.

(a) How high does the ball rise?
wrong check mark m

(b) How long does the ball remain in the air? (Hint: The time it takes the ball to rise equals the time it takes to fall.)

I dont know how to setup the equation.

Hint: the ball going up is fighting gravity and is decellerating at 9.8m/s^2 and stopping at 0m/s; then in accellerates at 9.8 m/s^2 until it reaches its initial speed at which it is caught. You are looking for TIME. If you can determine how long it takes to go up OR how long it takes to go down, you can determine how high it is.

Now read my tag line.

so do u multiply -9.8 m/s x 22.5?

Ok thank you for that question.

This one is confusing me too. It seems so easy yet I keep getting my answers wrong


Figure 5-26

http://i54.photobucket.com/albums/g94/arsenal4life3/5-26.gif

(a) t = 0 s and t = 5 s
wrong check mark m

(b) t = 5 s and t = 10 s
m

(c) t = 10 s and t = 15 s
m

(d) t = 0 s and t = 25 s
m


I only answersed the first one and put an answer of 30 m which was obviously too easy to be the answer. Im not sure how to do this one.

please help me on that question

If it is asking you to find the distance, find the area of the triangle (or the area "under the curve" for the given times.) So for 0<t<5, the area under the curve is 30*5/2=75m

Ok thank you for that question.

This one is confusing me too. It seems so easy yet I keep getting my answers wrong


Figure 5-26

http://i54.photobucket.com/albums/g94/arsenal4life3/5-26.gif

(a) t = 0 s and t = 5 s
wrong check mark m

(b) t = 5 s and t = 10 s
m

(c) t = 10 s and t = 15 s
m

(d) t = 0 s and t = 25 s
m


I only answersed the first one and put an answer of 30 m which was obviously too easy to be the answer. Im not sure how to do this one.30 is the velocity at the end of the interval. You're supposed to find the distance traveled during the interval. Remember that distance is the product of velocity and time. At the beginning of the time interval, velocity was 0. What was the average velocity?

"I see a failing grade in your future!"
Oh! I'm sorry! I though you said "psychic" homework! :o ;)

"I see a failing grade in your future!"
Oh! I'm sorry! I though you said "psychic" homework! :o ;)


:roflmao:

Serious question: when was the point of school completing tasks without understanding the subject?


See thread: Are Americans Dumb? for more information.

Never mind... you seem more interested in having someone do your homework for you than in learning physics. Good luck.

OH man. People with math questions should know to listen to Jschen.....

He only wants you to try and understand the problem. He will not tell you the answer or do it for you but will help you get through it, and understand it.

Maybe next time, we'll get a specific problem and targeted questions about the problem rather than a copy/paste job on an entire problem set. I'll be there then.

Ok thank you for that question.

This one is confusing me too. It seems so easy yet I keep getting my answers wrong


Figure 5-26

http://i54.photobucket.com/albums/g94/arsenal4life3/5-26.gif

(a) t = 0 s and t = 5 s
wrong check mark m

(b) t = 5 s and t = 10 s
m

(c) t = 10 s and t = 15 s
m

(d) t = 0 s and t = 25 s
m


I only answersed the first one and put an answer of 30 m which was obviously too easy to be the answer. Im not sure how to do this one.

The distance traveled between two points in time on a graph of velocity with respect to time is simply the area underneath the plot between those two points in time. So, just integrate between the two time points in question and there you go.