Homework Statement
On the construction site for a new skyscraper, a uniform beam of steel is suspended from one end. If the beam swings back and forth with a period of 1.10 s, what is its length?
Homework Equations
T=2*pi*sqroor (L/g)
The Attempt at a Solution...
Homework Statement
A hydrogen atom exists in an excited state for typically 5.79 ns. How many revolution would an electron make in the n=2 state before decaying?
Homework Equations
L=m*v*Rn=h(bar)*n
w=v/Rn
Rn=n^2*Ao
Ao=.053 nm
The Attempt at a Solution
Rn=4*.053=.212 nm...
Homework Statement
A proton is confined in a uranium nucleus of radius 7.41 fm. Determine the proton's minimum kinetic energy K less than or equal to delta K according to the uncertainty principle if the proton is well approximated by a Gaussian wave packet confined by the nuclear diameter...
Homework Statement
The wave function of a particle is Y(x,t) = A e^-kx*e^-iwt for x greater than or equal too 0, and it is zero everywhere else. What is the numerical value of the normalization constant A for k=5.55 1/nm and w =7.19 1/ps?
Homework Equations
intergral Y^2=1
The...
True Mentz144 but the Final KE does not equal the Initial KE. I finally figured it out. my KEi was right but I added masses when finding the new MoI not m*r^2 to the MoI of the disk.
the edge of the merry go round is moving at 11.336 m/s much fast then the man. So the merry-go-round slows down. I did conservation of momentum.
Initally...
Lmgr= I*w= 2310.58
Lp=I*(v/r)=623.326
Ltot= 2933.9
using this I found the angular speed of the merry-go-round fonal
2933.9/(Imgr...
Homework Statement
A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.686 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.99 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the...
Homework Statement
The crane shown in the drawing is lifting a 186-kg crate upward with an acceleration of 2.0 m/s2. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 130 kg. The cable is then wound onto a hollow cylindrical drum...
I did the following...
Mc=Mass canoe
Mp=Mass person
4.45= distance from shore+half distance of canoe
Cm=Cm
(Mc+Mp)*4.45/(Mc+Mp)=Cm
Cm=4.45m
4.45=3.4Mp+5.35Mc/(Mp+Mc)
4.45Mp+4.45Mc= 3.4Mp+5.35Mc
1.05Mp=.9Mc
1.1667Mp=Mc
Mc= 71.983kg
Is this correct? or am I not suppose to assume...
I thought of that a realized I don't know how to go about it being a center of mass problem. The only thing I really undestand is its the sum of the mass*d/total mass of the system. In this case I don't know what distances to use. Distance from the shore? Distance from the center of the...