Recent content by Kmol6

Homework Statement A 1.3m×1.3m×1.3m cube of nitrogen gas is at 20∘C and 1.5 atm. Estimate the number of molecules in the cube with a speed between 700 m/s and 1000 m/s. Homework Equations P= (1/3) (N/V) (Mv2rms)The Attempt at a Solution 20oc = 293 K 1.5 atm = 151950 Pa Solving for N N=...

Thank you!

1/2mv^2=qDeltaV? Then sub the answer for delta V into DeltaU=qDeltaV using q as 1.602X10^-19C and then plug Delta U into 1/2mv^2=DeltaU and solve for v^2 of the electron? (I think systematically, combining equations isn't easy for me) I got 2.2X10^6m/s ?

KEi + PEi = KEf+PEf 1/2mv^2 +mgh= 1/2mv^2 + mgh 1/2(9.11x10^-31kg)(51000)^2 + 0 = 1/2 (1.67X10^-27)(V)^2 +0 Vf=1284 m/s

Homework Statement A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 51000 m/s . What will be the final speed of an electron released from rest at the negative plate? Homework Equations...

Oh boy, thank you!

OK so the 165 should be -? 16938.5=1/2v^2 V= 184.05 m/s

As it is falling the KE is increased and PE is decreased

I've been questioning the height this whole time, as it is fired at a 45 degree angle the y component isn't just the height of the cliff, but that's the logic I'm going through. and if the height is in fact higher then 165m , how do I calculate it when I only have velocity? Also, I did notice...

using 1/2mv^2+mgh=1/2mv^2 + 0 The mass's cancel out, so if I fill in the following: 1/2 (175)^2 + (-9.8)(165) = 1/2 v ^2 ( my only question here is should I use the initial velocity or the horizontal velocity of 175 (sin45) = 123.7 m/s ?) 13695.5 = 1/2v^2 V = 165.5 m/s Is this correct...

The velocity of the x and y components are 175 (cos 45) = 123.7 and 175(sin 45) = 123.7 The ball is initially at 165m above the cliff, but projects up further ( i don't know how high though?) at that point the potential energy is equal to the final kinetic energy when the ball hits the ground...

Homework Statement A projectile is fired at an upward angle of 45 degrees from the top of a 165m Cliff with a speed of 175 m/s. what will be its speed when it strikes the ground below? (Use conservation of energy and neglect air resistance.) Homework Equations E= Ke +Pe 1/2 mv^2 + mgh =...

I definitely try and plug the numbers in as soon as I can, and I avoid combining equations. I don't trust my physics / math skills enough yet, it's been a decade since I've done any of this. o_O Thank you for the help and advice! I got the right number.

ok, If I use v^2 = vo^2 = 2a (x-x0) I think this is the one you're thinking of? ( a lot less work :P) 10.4^2=2(9.8)x x=5.52m which is close, but still not exact even with rounding? or I use conservation of energy and I get 1/2(.029)(510)=1/2(1.42)(10.4)^2 + (1.42)(9.8)h x= 4.98 m

Homework Statement A bullet is fired vertically into a 1.40 kg block of wood at rest directly above it. If the bullet has a mass of 29.0 g and a speed of 510 m/s, how high will the block rise after the bullet becomes embedded in it? Homework Equations 1. m1v1 +m2v2 = mfvf 2. x=xo +vot...