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Chanz
Questions (2)
1. A stationary curling stone is struck in a glancing collision by a second curling stone of equal
mass. If the first stone moves
4 answers
638 views
1. A stationary curling stone is struck in a glancing collision by a second curling stone of equal
mass. If the first stone moves
1 answer
1,133 views
Answers (826)
T = 2pi sqrt(m/k) Start the second at T/2
100 - mg sin30 = ma
tan-1(2/3) t = x/v = 85/3
mc(Tf - Ti)coffee + mc(Tf - Ti)milk = 0 solve for Tf (and you'll need to look up the specific heat of water)
Sure don't. After 1 sec both she and the bus are moving east at 2 m/s so someone on the ground sees 4m/s. At 2 sec the bus is moving 4m/s so someone outside sees 6m/s
Area = pi r^2 Look at your units and figure out what needs to be multiplied by what.
a) mv (gun) = mv (bullet) b) 1/2mv^2 = Fd solve for f (and watch your units)
1/p + 1/q = 1/f Solve for q (the height doesn't enter in to it)
yes
Square both sides, cross multiply, divide both sides by m^2, subtract 1 from each side, multiply both sides by -c^2
v^2 = 2gx Solve for x and don't forget to add in the initial 105.
Oooooh. Sorry. N
All I see to the right is an ad.
v^2 = 2gx solve for v
Definitely unit 9
23. They'll come to an equilibrium temp regardless.
Actually it can be greater than one. Velcro for example has a mu greater than one.
tan-1 (10/200)
See other post
Acceleration is not a force. The fourth force is the normal force the ramp exerts on the log.
KE = 1/2 mv^2 Solve for v I get 3.69
f = fo v/v-vs You'll have to look up the speed of sound at 10o, and don't forget to convert to m/s
Another conservation of momentum question. mv before = mv after. Solve for second v.
Conservation of momentum. Zero = zero.
Little pushy aren't you? Especially since we have no diagram.
Water at 900 is called steam and it can have a wide variety of densities based on pressure.
See next post
3.9 = vo cosθ (.555) ½ g (5552 ) = vo sinθ Divide eq 2 by eq 1 to clear vo do a tan-1 to solve
Nope. All I can tell you is that 100g of water has been displaced.
Not enough info
P = E/t
If you just do a FBD on the last cart and ignore all the rest a = F/m = 3.3/2.2 = 1.5 A classic example of useless smoke and mirrors.
You're in the wrong forum and you refer to a question that isn't there.
W = mgh Time has no bearing.
L = Iω You'll have to look up I for a solid cylinder and a mass at a distance. Also need to convert rev/sec into rad/sec. When you have a) and b) add them for your total L. Find the new I for masses at the new distance. Then conservation of L
v = k/r^2 solve for k Now use k and find v for r=1 and r=2. The difference of those (remember v is a vector) will be your answer.
You're on the right track. Note that x=0 when the rocket lands, solve for t and then x = vt to see how far it drifted.
vx stays constant at 16cos28 time to goal = 16.8/16cos28 vy at goal = = 16.8sin28(t)- 1/2gt^2 Then Pythagoean with vx and vy
τ t = ΔIω Solve for ω θ = ωt and s = rθ Same technique for part two with new τ
L = I omega you'll have to assume it's a solid sphere which is not quite correct (or you can look up the correct I for earth). Now how do you get omega? How many radians in how much time?
t = v sin35/g x = v cos35 t
t sound = x/vs t cannonball = x/vc difference is warning time
m1v1 = (m1+m2)vf 1/2 (m1 + m2) vf^2 = 1/2 kx^2 solve for x
time to get to 8th floor: t = v/g distance fallen during that time: y = 1/2gt^2 divide by 8m/floor and yes it fell 15 floors total (watch, the last is 12m)
mu(m1+m2)g = (m1+m2)a so mu(g) = a speed before slide begins: v = sqrt(2ax) m1 vi = (m1+m2)v solve for vi
P = 2 pi sqrt(L/g) Solve for L L - Lcos5.34 = h now mgh = 1/2mv^2 and the m's cancel, solve for v
P = W/t
v = f(lambda) v in this case is x/t then solve for f
v = f (lambda) Solve for lambda
Normal: Fn=mr(omega^2) You'll need to change rev/sec into rad/sec Friction: Ff= mg Coefficient of friction mu = Ff/Fn
