Physics 20 - help

1. Jan 8, 2004

Pepsi

I was given a few questions from my teacher to practice for my final exam which is coming up soon, here are the ones I am having troubles with, if anyone can show me how to do em that would be great.

1. Astronauts release two 10.0 kg objects from a space shuttle and place them 10.0m apart, at a location where no other gravitational forces act. The objects are initially at rest. Using Newton's laws and showing your method of calculation, determine the time (h) it will take the gravitational attraction between the two objects to move them 1.00 cm closer together.

and then this one is very hard to explain because there is a picture... its of truck, I'll do my best to explain it.

Use the diagram that shows a truck travelling at a constant speed a long a curve in the road to answer the question.

alright let me explain this... the 1 is a force moving up from the top of the truck.

2 is the force on the tire that is moving ->
3 is the force on the tire moving down
4 is the force on the truck moveing out from its sides
and 5 is the force come from the back of the truck

By number list the forces that are being exerted on the truck and indicate their source.

I have a few more questions but i dont want to hastle anyone, thanks in advance.

2. Jan 8, 2004

Integral

Staff Emeritus
Fisrt of all we have a homework help forum, in the future please use it. Be sure to read the "read me first" sticky at the top of the forum.

1. You have been given an expression for the force of gravity between two bodies use that and Newtons F=ma to determine an acceleration.

Can you take it from there?

2. Sorry, it is not clear what you are looking for on this one.

3. Jan 8, 2004

Pepsi

so would i do...

fg=Gm1m2/r^2

fg=6.67x10^-11

then take that... and do f=ma? so 6.67x10^-11/10kg, or would I do 20kg, or am I way off track?

4. Jan 8, 2004

Jimmy

You have to calculate the acceleration of each mass separately and then add those values to get the total relative acceleration. The gravitational force that each mass is exerting on each other is the same. If you used 20kg in the formula a=F/m, you would end up with the wrong answer. However, since both objects have the same mass, you can calculate the acceleration for one and then double that value. That would be the relative acceleration which can be used to calculate the time in your problem.

Side note: As the objects move closer, the acceleration will increase. However, since you only need to calculate the time in hours that the objects will move 1cm closer, you can consider the acceleration constant. The difference of the gravitational force between the objects at 10m and 9.99m is not going to be significant when considering how many hours it will take to move across that distance.

Last edited: Jan 9, 2004
5. Jan 8, 2004

Pepsi

wow thanks jimmy

so i go...

f=ma..... 6.67x10^-11/10kg... then times that by 2... and I get... 1.334.10^-11.

and then i use that as my acceleration... so a=v/t

somehow i got 7.488 x10^-11 as my final answer does this look right?

6. Jan 8, 2004

Jimmy

Everything looks good up until the a=v/t. Remember, you have to figure the time it takes an object to cover a distance of .01m at an acceleration of 1.334.10^-11 m/s^2. Velocity has never been mentioned.

Try to remember a formula which equates time, distance and acceleration.

7. Jan 8, 2004

Jimmy

Hint:

$$V = \frac{2D}{T}$$

and...

$$V = A T$$

If A=B and A=C, then B=C...

Last edited: Jan 8, 2004
8. Oct 17, 2010

boosgore

I need help with this question, I'm not sure of which angle I need to find

An airplane, flying with a velocity of 900km/h [120degrees], experiences a 50km/h wind blowin at 190degrees

a) to reach a destination that is 500km away at 120degrees, the pilot must head the airplane in a direction of ???degrees, expressed using the cartesian method.