Lenses/Images -- Find final image
Two converging lenses, each having a focal length equal to 8 cm, are separated by 35 cm. An object is 30 cm to the left of the first lens.
(a) Find the position of the final image using both a ray diagram and the thin-lens equation
(c) What is the overall...
Heat/First Law of Thermodynamics/Calorimetry -- Have work done
A calorimeter of negligible mass contains 920 g of water at 303 K and 48 g of ice at 273 K. Find the final temperature T.
Solve the same problem if the mass of ice is 460 g.
Formula Ive used
Mice * Lf + Mice * Cwater *...
The period of an oscillating particle is 56 s, and its amplitude is 18 cm. At t = 0, it is at its equilibrium position. Find the distances traveled during these intervals.
(a) t = 0 to t = 14 s
(b) t = 14 s to t = 28 s
(c) t = 0 to t = 7 s
(d) t = 7 s to t = 14 s
Water flows at 0.67 m/s through a 3.0 cm diameter hose that terminates in a 0.35 cm diameter nozzle. Assume laminar non-viscous steady-state flow.
(a) At what speed does the water pass through the nozzle?
(b) If the pump at one end of the hose and the nozzle at the other end are...
Fluids -- pressure
Blood plasma flows from a bag through a tube into a patient's vein, where the blood pressure is 10 mmHg. The specific gravity of blood plasma at 37°C is 1.03. What is the minimum elevation the bag must have so the plasma flows into the vein?
Prof gave this problem for us...
well the textbook has the same problem explained but different velocity and it uses that equations and same steps I posted. But I don't know what I did wrong mathematically to give me a negative height. I was wondering if someone can work it out and see if they get the same answer or diff.
This is the final equation:
1/radius_f = -Vi^2/2*G*M_e + 1/R_e
then I solve for radius_f then I divide it by 1
then I subtract it from the earth's radius to get the height but it gives me a negative height.
Gravity -- maximum height
An object is projected upward from the surface of the Earth with an initial speed of 3.7 km/s. Find the maximum height it reaches.
Ive used the Energy conservation
Kf + Uf = Ki + Ui
1/2 mvf^2 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/ri
= 0 - G*M_e*m/rf = 1/2 mvi^2 -...
Yes, it makes sense, thanks for explaining.
And I got what r is now that you related it as a circle.
For part C:
What you said earlier
time = distance/speed
where distance is the orbits circumference/distance covered
It would be 2Pi*(86.37 x 10^6) / 2.15km
I get a big answer of 252528.6976s...
Ok, I got the mass of the earth 5.98 × 10^24 kilograms from the textbook.
So the raidus of the Earth is 6.37 x 10^6m which my textbook says.
So I add 8.00 x 10^7 m to it? So is 86,370,000 which is 86.37 x 10^6
gravitational Force = G*m1*m2 / r^2
G = 6.67 x 10^-11 Nm^2/kg^2
F = (6.67 x...
Gravity Problem -- satellite
A satellite with a mass of 250 kg moves in a circular orbit 8.00 x 10^7 m above the Earth's surface.
(a) What is the gravitational force on the satellite?
(b) What is the speed of the satellite?
(c) What is the period of the satellite?
An 80cm long pendulum with a .60kg bob is released from rest at an initial angle of Theta with the vertical. At the bottom of the swing, the speed of the bob is 2.8m/s.
a)what is Theta?
b)What angle does the pendulum makes with the vertical when the speed of the bob is 1.4m/s?
A 2.4kg block is dropped onto a spring with a force constant of 3.96*10^3 N/m from a height of 5m. When the block is momentarily at rest, the spring is compressed by 25cm. Find the speed of the block when the compression of the spring is 15cm.
External Work = change in...
A force Fx acts on a particle that has a mass of 1.5kg. The force is related to the position x of the particle by the formula Fx = Cx^3, where C = .50 if x is in meters and Fx is in newtons.
a) What are the SI units of C?
b)Find the work done by this force as the particle moves from x=3.0m...
Hi, alphysicist, thanks for responding.
The collision itself, the angular momentum is conserved.
Angular momentum = I * omega
Do I consider the before collision or after collision first?
Do I use conservation of energy on the particle or the rod?
Im just confuse on how to start this.
A uniform rod of length L1 = 2.2 m and mass M = 2.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1.8 m from the hinge. The particle sticks...
Oh, your right, Im solving for initial angular speed. So would Kf be 0? Since all we care if it gets barely over?
Ui + Ki = Uf + kf
MgR + 1/2(2MR^2)*omega initial ^2 = 2MgR + 0
Also Doc, when would I use the parallel axis theorem on problems? Like how would I know I should use it?
So Parallel Axis Theorem is I = Icm + Mh^2
I = MR^2 + MR^2
I = 2MR^2
So I can still apply the energy of conservation equation?
Ui + Ki = Uf + Kf
MgR + 0 = 0 + 1/2 (2MR^2)*omega ^2
MgR = 1/2 (2MR^2)*omega ^2
omega = square root of g/R
this is the maximum angular velocity
root of 9.81m/s/1.1m =...
I was doing a same problem but with different numbers so please ignore the R = .75m, suppose to be R = 1.1m
I went to tutoring and this is what the tutor did, I didn't really understood it.
Parallel Axis Theorem is I = Icm + Mh^2
Im not sure what to do.
Rotation -- Anuglar speed
A uniform ring 2.2 M in diameter is pivoted at one point on its perimeter so that it is free to rotate about a horizontal axis. Initially, the line joining the support point and center is horizontal.
(a) If the ring is released from rest, what is its maximum angular...
It says it gave the wheel an angular velocity of +610 rev/min
to change this to rad/sec.
I got 63.87905062 approx 64 rad/sec.
63.87905062 rad/sec / 24s = 2.661627109 approx 2.7 rad/s^2 is the acceleration
Tnet = I alpha
so Tnet/alpha = I
40 Nm /2.661627109 rad/s^2 = 15.02840110 approx 15 is the...