Stumped with some basic physics

D

daunt

Veteran X
Never thought I'd bring my homework here, but my teacher has been utterly useless in helping review for the final on Thursday. So yeah, here's a couple basic high school physics problems that shouldn't take too long if you know what you're doing. I already did 20 something on my own so it's not like I'm one of those lazy bastards:

A rock is dropped into a deep cavern. 3.8 seconds later the splash is heard. If the speed of the returning splash sound is 330 m/sec, how deep is the cavern?

-I came up with 63.5 m but I've been told I'm off by a hair. I don't know the proper setup for the problem.

A 6 lb force pushes a 20 lb weight down a 19 degree frictionless slope. What is the acceleration of the 20 lb weight?

-Don't know how to set the problem up and the standard units don't help any.
 
im no physics major or wiz and hate the course
but
a rock dropped would fall at 9.8 m/s
it falls at most 3.8 seconds.
9.8m/s x 3.8s = 29.4m
the speed of the returning sound is about 11 times the distance you have, so somewhat redundant. Although I suck at physics .... Would say its smaller then 29.4
 
BLiSTeD said:
im no physics major or wiz and hate the course
but
a rock dropped would fall at 9.8 m/s
it falls at most 3.8 seconds.
9.8m/s x 3.8s = 29.4m
the speed of the returning sound is about 11 times the distance you have, so somewhat redundant. Although I suck at physics .... Would say its smaller then 29.4
Click to expand...


first off this is wrong. The rock accelerates at 9.8 m/s^2, thats not its velocity.

ill work out the rest and edit the post
 
god dan
you are so fucking stupid :lol:
j/k :heart: ya

convert lbs to N...easiest IMO

you have the 6lb force pushing down on the block + the mg ( cos 19 deg)

so that sum = F on the block
F=ma, divide by 20, you get a.

i'm not too sure about the sin vs cos...i THINK it's cos...it could be sine tho.
 
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I thought senioritis was some sort of dumb excuse to be lazy till I started getting steadily dumber about three months ago. I've just totally spaced on a bunch of stuff I learned earlier this year.
 
daunt said:
A rock is dropped into a deep cavern. 3.8 seconds later the splash is heard. If the speed of the returning splash sound is 330 m/sec, how deep is the cavern?
Click to expand...

330m/s X 3.8 sec = distance

distance= v0t+ .5at^2
v0 (pronounced v knot) = initial velocity = 0
so distance = .5at^2

is that it?
 
what in the fuuuuck, how are you in physics 2 and you think things travel at a constant velocity of 9.8m/s while falling??
 
cancer said:
330m/s X 3.8 sec = distance
Click to expand...
that's not right because it's 3.8 seconds from the time that you dropped the rock, so a certain portion of the time was the time it took for the rock to drop (X) and the time it took fro the sound to come back (3.8 - X)
 
daunt said:
Never thought I'd bring my homework here, but my teacher has been utterly useless in helping review for the final on Thursday. So yeah, here's a couple basic high school physics problems that shouldn't take too long if you know what you're doing. I already did 20 something on my own so it's not like I'm one of those lazy bastards:

A rock is dropped into a deep cavern. 3.8 seconds later the splash is heard. If the speed of the returning splash sound is 330 m/sec, how deep is the cavern?

-I came up with 63.5 m but I've been told I'm off by a hair. I don't know the proper setup for the problem.

A 6 lb force pushes a 20 lb weight down a 19 degree frictionless slope. What is the acceleration of the 20 lb weight?

-Don't know how to set the problem up and the standard units don't help any.
Click to expand...

problem 1:

y = y0 + v0t + a/2t^2

y = 0 + 0t + g/2 * 3.8^2

y = 9.8/2 * 3.8^2

y = 70.756m

problem 2:

sum(f) = ma

write out the components
x: mgcos19 + F = ma
y: mgsin19 = ma

y isn't a factor since it's frictionless and you establish a coordinate system with x parallel to the plane

x: mgcos19 + F = ma

solve for a

a = (mgcos19 + F)/m

m = 20
f = 6

so:

a = (20(9.8)cos19 + 6)/20

a = 9.57 m/s^2


check my math....
 
daunt said:
I thought senioritis was some sort of dumb excuse to be lazy till I started getting steadily dumber about three months ago. I've just totally spaced on a bunch of stuff I learned earlier this year.
Click to expand...
I would help, but I forgot all of my physics for this very reason.
 
because my phy I class all sucked
therefore our curve per test was really low
i got at 4 right on all tests therefore getting a C on each test and getting a C in phy I.
 
invar said:
that's not right because it's 3.8 seconds from the time that you dropped the rock, so a certain portion of the time was the time it took for the rock to drop (X) and the time it took fro the sound to come back (3.8 - X)
Click to expand...

yeah i just saw that
my bad
i read it too fast
 
BlodBath-VuP- said:
problem 1:

y = y0 + v0t + a/2t^2

y = 0 + 0t + g/2 * 3.8^2

y = 9.8/2 * 3.8^2

y = 70.756m

problem 2:

sum(f) = ma

write out the components
x: mgcos19 + F = ma
y: mgsin19 = ma

y isn't a factor since it's frictionless and you establish a coordinate system with x parallel to the plane

x: mgcos19 + F = ma

solve for a

a = (mgcos19 + F)/m

m = 20
f = 6

so:

a = (20(9.8)cos19 + 6)/20

a = 9.57 m/s^2


check my math....
Click to expand...

I'm looking this over now. Thank you much.

*edit* I don't think the time it takes for the sound of the splash to come back is acocunted for in the first problem.

Second problem is in standard units rather than metrics. I think I can extract the concept though because that looks right.
 
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daunt said:
I'm looking this over now. Thank you much.

*edit* I don't think the time it takes for the sound of the splash to come back is acocunted for in the first problem.

Second problem is in standard units rather than metrics. I think I can extract the concept though because that looks right.
Click to expand...

hopefully.

Ive been trying to work this problem out for the exact time and i got a headache. physics skillz aint what they used to be. I got stuck but heres what i had:

so. We have to account for the time of the rock dropping and the sound wave accending.

d= di + vi(t) + 1/2(at^2)
vf = vi + at

x = time for sound to ascend


for wave:
0 = -d + 330(x) + 0
d = 330x
x = d/330

for rock:
d = 0 + vi(3.8-x) + 1/2(9.8(3.8-x)^2)
plug in x

d = 0 + 0(3.8-d/330) + 1/2(9.8(3.8-d/330)^2)
d = 4.9(3.8-d/330)^2)
d = 4.9 ((14.44-d^2)/108900)
1 = (70.756/d - 4.9d)
1= 6.49734x10^-4 /d - 4.4995x10^-5d

^AHHH
 
cocaine said:
1= 6.49734x10^-4 /d - 4.4995x10^-5d

^AHHH
Click to expand...

I read nothing but your last line so I'm not sure if it's right, but just move the third term to the left side, multiply through by d, and use the quadratic formula.

4.4995x10^-5d^2 + d - 6.49734x10^-4 = 0
 
d = distance
v = velocity when rock impacts
t = time for rock to hit
3.8 - t = time for sound to travel back
330 = speed of sound

9.8 * t = v

distance travelled is average velocity * times. we have represented v, so we'll use it now. v/2 = average velocity since we know starting velocity is 0. it spends an equal amount of time at both extremes, so dividing by 2 is allowed.

d = vt/2

(3.8 - t) * 330 = vt/2

1254-330t = vt/2
2508-660t=vt
2508-660t=9.8t^2

9.8t^2+660t-2508 = 0 (punch into POLY for TI-85 since it is a quadratic equation)

t=3.6068324881

plug t back into above (9.8 * t = v)

v=35.3469583834

plug v and t into above again (d = vt/2)

d=63.7452789264
 
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