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Law of sines: c/sin(A) = c/sin(B) = c/sin(C) Do I have to solve this as a system of equations to solve for the sides? But where does the given perimeter come into play?

Homework Statement All three sides of an isoceles triangle are given along with its perimeter. Find the length of each side. A=97.433 B=41.283 C=41.283 Perimeter=24.78in Homework Equations p = a+b+c a2=b2+c2 - 2*b*c*cos(A) The Attempt at a Solution Would you somehow find the...

I got it. Thank you!

Homework Statement The Ko meson is a subatomic particle of rest mass MK = 498 MeV/c2 that decays into two charged pions, Ko \rightarrow \pipositive+ \pinegative. Both pions have the same mass, m\pi = 140 Mev/c2. A Ko at rest decays into two pions. Use conservation of energy and momentum to...

Homework Statement A U-shaped conducting rail that is oriented vertically in a horizontal magnetic field. The rail has no electric resistance and does not move. A slide wire with mass m = 10 g, L = 0.45 m, and resistance R = 0.10 Ω can slide up and down without friction while maintaining...

Homework Statement An electron in a cathode ray tube is accelerated through a potential difference of ΔV = 11 kV, then passes through the d = 4 cm wide region of uniform magnetic field. What field strength (in mT) will deflect the electron by 10(degrees)? (Hint: is it a reasonable...

Before I submit my answer, would you agree with using conservation of energy and the answer that I got?

So, now at 90% of 172,800 J = 155,520 J Using conservation of energy: 155,520 J = mgh 155,520 J = (8.1 kg)*(9.8 m/s^2)*(h) h = 1959.18 m ? This seems like a large value but reasonable because the mass is not that great.

Homework Statement A fully energized 12V battery is rated as "4 Ampere-hours." Suppose this battery is connected to a motor that is 90% efficient at converting electrical to mechanical power. How high could this battery-supplied motor lift a 8.1 kg mass? Homework Equations V = I*R P =...

Kinetic + potential = Emech (1/2m(Vi^2)) - q(E)*(Xi) = (1/2m(Vf^2)) - q(E)*(Xf) Then we changed the equation to this: qE(20cm) = 1/2(mass of electron)*((Vf^2) + (Vi^2)) Then we solved for Vf: Vf = 3.61697 x10^7 m/s

So the force would be F = (50000 V)*(1.602 x10^-19 C) F = 8.01 x10^-15 N

Homework Statement An electron with an initial velocity of 4.7 x 10^7 m/s in the x-direction moves along a trajectory that takes it directly between the center of two 20 cm x 20 cm plates separated by 1 cm. The plates are connected to a high voltage power supply so that the potential...

Thanks for your help, foxjwill!

When I integrate the field of the semicircle, I come up with a field of magnitude -147,858 N/C...?

we haven't even started studying gauss law yet...

[SOLVED] E-field at the center of a semicircle Homework Statement Calculate E at center of semicircle: A uniformly charged insulating rod of length L = 12 cm is bent into the shape of a semicircle as shown. The rod has a total charge of Q = -6 nC. Find the magnitude of the electric...

I used the wrong radius in the previous post. I used .03 and got the right answer. Thanks for all your help Doc Al!

Ok. That 1.60535 is correct. For part b now, would i just use V = Rw to find the angular speed. So, w = 1.60535 / sqrt(.02^2 + .03^2) w = 44.5238 and this would be in rad/sec?

Now I'm getting 1.60535 m/s.

(.43Kg)*(9.8m/s^2)*(-.7m) = -2.9498 I'm getting the same number from before.

that number is the M*g*h for the hanging mass which should be right. did I forget to include friction on the right side of the equation?

I recalculated the -.999761 and got -.914667. Then I got 1.79716 m/s and it's still wrong.

1/2(M1*Vi^2) + 1/2(M2*Vi^2) - friction + 1/2 (I)(V/R)^2 = 1/2(M1*Vf^2) + 1/2(M2*Vf^2) + M2*g*h + 1/2 (I)(V/R)^2 I found (I) to equal .000228 1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) + 1/2(.000228)(.82/.03)^2 = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) +...

Now I'm getting 1.75919 m/s. This seems way to close to my answer from post #3.

I added 1/2 Iw^2 to each side using the initail veloctiy for w on the left side and w on the right side is the same as the Vf on the right side? When I do this I'm getting 2.397m/s which wouldn't be in the 10%.

When I use a negative height, I get 1.75888 m/s and it tells me my answer is wrong but within 10% of the correct value.

[SOLVED] Speed of a Block Homework Statement The sliding block has a mass of 0.830 kg, the counterweight has a mass of 0.430 kg, and the pulley is a hollow cylinder with a mass of 0.350 kg, an inner radius of 0.020 m, and an outer radius of 0.030 m. The coefficient of kinetic friction...

Nevermind...I figured it out.

[SOLVED] Finding friction from tangential acceleration Homework Statement A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 2.25 m/s^2. The car makes it one quarter of the way around the circle before it skids off the...

Homework Statement A m1 = 45.0 kg block and a m2 = 97.0 kg block are connected by a string as in the figure below. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 45.0 kg block and incline is 0.250. Determine the change in the kinetic energy...

[SOLVED] Object Dropped on a Spring Homework Statement A 1.30 kg object is held 1.20 m above a relaxed, massless vertical spring with a force constant of 310 N/m. The object is dropped onto the spring. (a) How far does the object compress the spring? ________m (b) How far does the...

To find part b which is the work done by the gravitational force from point a to c would I use... mgHa - mgHc = Kinetic energy = Work ?

My first answer was wrong cause i solved the formula wrong but i fixed it and found Vf = sqrt(2(gHa - gHb)) which ended up to be 6.261 m/s.

so using mgHa = mgHb + 1/2mVf^2, i got 5.77m/s at point B.

Since Vo = 0m/s wouldn't the 1/2mv^2 on the left side of the equation cancel and then I would just solve for my Vf on the right side

There is a picture attached but if you can't see it, point B is 3.2m and point C is 2m.

[SOLVED] Particle movement down a track Homework Statement A particle of mass m = 6.60 kg is released from point A and slides on the frictionless track shown in the figure below. (ha = 5.20 m.) (a) Determine the particle's speed at points B and C. point B____________ m/s point C...

That would explain my mistake. I was using the horizontal component of the velocity. When I use 3.13 I get 1.03827 sec, which is correct. Thanks a lot learningphysics!

Homework Statement A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the surface of a table, at the top of a θ = 35.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m. How much time has elapsed between when the block is...

Alright, I get it now. Thanks for your help Astronuc and Doc Al!

so, the horizontal component of 761.491 N would be sin(5)*761.491 which equals 66.3683 N

I entered 9.764 N for the horizontal component of the tension force and it said it was incorrect.

Ok, I found the acceleration to be .851m/s^2. Now, to find the horizontal component of the force i would set that acceleration equal to T*Sin(5)? Tsin(5) = a T = .851 / sin(5) T = 9.764 N ?

Homework Statement Consider a conical pendulum with a 78.0 kg bob on a 10.0 m wire making an angle of θ = 5.00° with the vertical. (Consider positive i to be towards the center of the circular path.) (a) Determine the horizontal and vertical components of the force exerted by the wire on...

There is no option in my "thread tools" to mark problems as solved. That's where the option should be right...?

Ok. The answer was correct. Thanks Doc Al!

sin(angle) = v^2 / gR ?

so to find the velocity i would take the square root of g*R? v = sqrt(9.8*.385) v = 1.9424 m/s then, (v/(2pi*R)) * 60 = revs/min (1.9424/(2pi*.385))*60 = 48.178 revs/min

So I set the normal force equal to zero just before it would leave the track and use.. normal - mg = m(v^2)/r masses would cancel. 0 + g = (v^2)/r gr = (v^2) sqrt(gr) = v when i take the sqare root of 9.8*15 I'm getting 12.1244m/s which sounds reasonable.

Cos(62.5) = (v^2)/gR That was the equation that our professor lead us up to in class to find velocity. Wouldn't the component of the weight in the radial direction just equal mgCos(62.5)?