Some physics problems i'm stuck on

[gaspacho soup]

Hey all, i have a big test this monday and i've been doing some practice problems. But their are 6 probs (in the motion section) that i can't do, can someone give me a hand?

1) Express 1 gram per cubic centimetre (g cm-) in SI units.

A) 1/100kg m^(-3)
B) 1000g m^(-3)
C) 1000kg m^(-3)
D) 1.0 x 10^6g m^(-3)

This question seems to be fairly straightforward, and i worked out a figure of 0.1kg m^(-3). The only trouble is that it's not listed in the answers. I think i mighta incorrectly assumed that 100 cm^(-3) = 1 m^(-3).

2) Calculate the component of. (i) vector a in the direction of b; (ii) vector b in the direction of a (iii) Vector a in the direction perpendicular to b

|a| = 3 units, |b| = 3 units, theta = 60 degrees.

http://homepage.mac.com/creacha/.Pictures/gaspacho1.jpg


3) A baloon has been ascending vertically from the ground at a constant speed for 10.0 seconds when the stone is let fall from the base of the balloon. If it takes the stone a further 5.0 seconds to reach the ground, find

a) The speed of the baloon, and
b) the height of it's base at the instant when the stone is released


http://homepage.mac.com/creacha/.Pictures/gaspacho2.jpg

I spent about 30min on this problem with no luck


4) A 30kg block is pusehd from rest up a plane, which is inclined at 30 degrees to the horizontal, by a force of 500N. The coeffiecient of sliding riction is 0.25. Find

a) The acceleration
b) The velocity of the block after it has moved 10m along the plane.


For this problem i used F(friction) = uN = 0.25 * mg = 0.25 * 30 * 9.8. I did it incorrectly i'm guessing.



4) Find the sum of the three forces shown. a) graphically b) anayltically.

http://homepage.mac.com/creacha/.Pictures/gaspacho3.jpg

Now for this problem, i thought i got the answer correct, but it doesn't match the answer given. So i need someone to confirm that my answer was right ^_^


5) Is the same time necssary for a cutter to travel a distance of 1 Ian upstream and back on a river (current velocity is equal to 2kmph) as on a lake (in still water) if the velocity of the cutter relative to the water is, both cases, 8kmph? Find the ratio of the first time to the second time.


I had no idea what they meant by 'Ian'.

[pog21]

Not sure what this has to do with flash.
Maybe why no replies...
1) = c.
To get cm to m times by 100. So for 3 dimensions, times by 1,000,000. Now divide by 1000 cos you're going from g to kg.

[gaspacho soup]

lol...sorry..i just looked up 'physics message boards' on google...:P

[john_bws]

2. dot product / projections and pythogrean theorm

cos 60 = .5

a proj b = ||a|| cos 60 = 1.5
b proj a = ||b|| cos 60 = 1.5
a perp b = (1.5*1.5 + 3*3)^1/2 = 3.35

3. if you know how long it takes something to fall you know how high it was dropped from

s = 1/2(9.8)(5)^2
s = 122.5

that takes care of b. speed of the baloon can be solve by knowing it took 10 secs to get to 122.5 height.

122.5 = v(10)
v = 12.25

that takes care of a. you dont always have to go in order.


4. three words. free body diagram. i cant really do one here but i will tell you have you 4 forces total. force due to gravity in the x- and y-direction (because it is on a slant), force of 500N pussing along the ramp, force of friction in the opposite direction

               /.\  Fn
                |
                |
          |-----|-------|
Fg sin30  |     |       |
    <-----|     |       |-----> 500 N
          |     |       |
    <-----|-----|-------|
Fg cos30 mu     |
                |  Fg cos30
               \./

sum the forces

Fg = 30(9.8) = 294
Fg sin30 = 147
Fg cos30 = 254.61
Fg cos mu = 254.61(.25) = 63.65

sum Fx = 500 - 147 - 254.61 - 63.65 = 34.74

newton's 2nd law

sum Fx = ma --> 34.74 = 30 a

a = 1.158

and there you go

the second #4. to do it analytically break the forces into their components and sum the components

result Fx = 25 cos 225 + 60 cos 330
result Fy = 25 sin 225 - 11 + 600 sin 330

go get em

[eMenendez]

Remember that the stone dropped from the ballon has an initial velocity in the +y direction. Use the constant acceleration equation:

xf = xi + vt + .5at^2

0 = xi + vi(5) + .5(-9.8)(25)

Since the ballon has been rising at a const. speed for 10 secs, xi in the above equation is equal to: vi(10).

Substitute in and solve for vi, then go back and find xi.