Chanz

h = 2830sin14.6 = 713.4 mgh = 75*9.8*713.4 = 5.24e5 joules

9 years ago

In y to find initial v 20 = vi(4) +.5(-9.8)(4sqr) so vi = 24.6 In y to find final v vf = vi + at = 24.6 -(9.8*4) = -14.6 In x since angle = 60 vf = 14.6/tan60 = 8.4 In x vi=vf = 8.4 x = vi t = 8.4 * 4 = 33.7 Using pythagorean with initial x and y values v...

9 years ago

Phi = EA = [kq/r(sqr)] * [4(pi) r(sqr)] = 4 kq(pi) = 4 * 9e9 * 5e-6 * 3.141 = 5.7e5 Note r cancels

9 years ago

E1 = kq/a1(sqr) at angle of 60 E2 = kq/a2(sqr) at angle of 120 x components cancel y components sum = 2(cos30)kq/a(sqr)

9 years ago

If we place the -2.5 charge at zero and 6.0 charge at 1 the E field will be zero somewhere (distance x) on the negative x axis. so k q1/x(sqr) = k q2/(x+1)sqr or q2/q1 = [(x+1)sqr]/x(sqr)= (x+1/x)sqr so 2.4 = (x+1/x)sqr 1.55 = x+1/x = 1 + 1/x .55 = 1/x x...

9 years ago

Since mass is irrelevant for part b I'm assuming they want the force of the push in part c which will be 584/1.92 (Fd =mgh)

9 years ago

Area under an F vs t curve is momentum. F is linear so we get a triangle: .5 (58) 25 = 10 v v = 72.5

9 years ago

I'm lost on your notation but... Position is simply (3t^2 + 7.2t + 28)i +(.28t^2 - 9.1t +30)j v = dx/dt: (6t + 7.2)i + (.56t - 9.1)j Magnitude of WHICH angle? tan-1(j component @ 20 / i component @ 20)

9 years ago

x1 = .5 (9.89).52^2 = 1.32 x2 = .5 (9.8).27^2 = .36 Distance between is .96 and center of mass 1/3 distance from stone 2 = .31. COM is therefore .36 + .31 = .67 b. v^2 = 2ax = 2(9.8).67 = 13.1

9 years ago

b. 1.58e9/33 = 4.8e7 kg coal for power. But only 33% efficient so total coal = 4.8e7 +(1.5*4.8e7) = 1.2e8 kg coal. Add 2.8e8 kg oxygen. Total reagents = 4.0e8

9 years ago

(a) sum of torques = zero so mg(10 cos(theta) = Fc(20 sin(theta)) (mass is assumed at center) sum of forces = zero so Fay = mg (really the normal component of Fa). and Fc = Fax = (mu)Fay So Fc = (mu)mg (if you assume the 18kg given in part b = .5*18*9.8)...

9 years ago

Moment of Inertia depends on the shape of the object. Surface tension depends on the liquid and its temperature.

9 years ago

t to fall is sqrt(2x/g) = .56 distance car moves = .55*22 = 12.17 rail car moves .55*(8.7)= 4.78 tot = 12.17+4.78

9 years ago

delta U = Q - W and W in this case = P delta V. Use standard pressure, change in V is given and heat added is given.

9 years ago

I think bob meant twice the stretch. 30cm

9 years ago

1. Momentum before = after In x net is 1.25cos44 - .92 cos71 = .6 in positive x In y net is .92sin71 + 1.25sin44 = 1.74 positive y. Use pythagorean and tan-1 to find magnitude and direction 2.Ft = delta mv = .62(2-1.7) Divide t to get F. 3. Make left posi...

9 years ago

First find temp at the join. Since Q must be the same k1 A deltaT t/L = k2 A deltaT t/L. A, L and t are the same so k1 (100-T) =k2 (T-0) and T is about 65C. Now go back and use Q =kA deltaT t/L

9 years ago

a) Presumably zero. All forces should cancel. b) The four matched corners will again cancel so this leaves only the unmatched corner.the distance between them can be found by pythagorean with one side 4nm the other (a diagonal across the bottom say) is 4s...

9 years ago

Unfortunately two sides do not determine a triangle. You're going to need an angle or the third side to do this one.

9 years ago

One electron is 1.67e-19 C. Divide .6e-9 by 1.67e-19 for your answer. b) Co has an atomic number of 27 and an atomis mass of about 59. So 41 grams is about 41/59 moles. Multiply by Avagadros (6.02e23) to get total number of electrons. The get the ratio fo...

9 years ago

Maybe I'm oversimplifying but this looks like a simple Coulomb's Law with q1 = 2e-29, q2 = 1.67e-19 and r = 25nm.

9 years ago

This would be a pretty easy superposition problem if we knew which charge was at which corner. Let's assume and you can make the adjustments to fit your diagram. I'll put q1 at the origin, q2 upper left, q3 upper right, and q4 lower left on the x-axis. Fi...

9 years ago

Steve, think you're a bit off on the height at the end of the burn. SB 331m. And I would subtract out g, the problem gives a flat 34.4 so v at that time is 151m/s. In the second stage v goes from 151 to zero under influence of gravity so y = v^2/2a = 151^...

9 years ago

Sorry would NOT subtract out g in part 1

9 years ago

Or if your Physics teacher is picky vertical is 9.8 * 2 = 19.6

9 years ago

Ack! Careful. This is not an energy problem, it's a Hooke's Law force problem. F =mg = 2kx k = mg/2x = 3.2(9.8)/(2*.32)

9 years ago

You're correct Lilly. It's not a very good question anyway. It will fall in 5.7 seconds no matter how fast the thing is going. Who knows whether they get the supplies or not.

9 years ago

Remember. The proton is being pushed by the plus charge and pull by the negative. If you put it between the two charges it would be draw right by the positive and pushed right by the negative. If we put it beyond the second charge it's pulled left by a mu...

9 years ago

Sure. Put it in a vacuum and ambient temperature is usually high enough to make it boil. It's not hot, but it boils.

9 years ago

1/f = 1/p + 1/q f = 10, p = 5. They value and sign of q will clue you into its nature. Ohh. And there's no need to shout :)

9 years ago

1. A stationary curling stone is struck in a glancing collision by a second curling stone of equal mass. If the first stone moves away at a velocity of 0.92 m/s [N71oW] and the second stone moves away at a velocity of 1.25 m/s [N44oE], what was the initia...

9 years ago

bob, can you show me your perspective on how to approach these questions pls? Each question.

9 years ago

F = mg sin30/.75 Although I've never heard of efficiency of an inclined plane.

9 years ago

For question one you don't need energy. Both stones are defined after with magnitude and direction. All of that momentum had to come from the second stone as the first was at rest. It's a simple addition of two momentum vectors. Question 2 is indeed a vec...

9 years ago

Too much work. ALL of the momentum must go to the second ball for perfectly elastic involving same masses(should be a give away that no info is provided post-collision). FT = mv 1.45/.145 = v You can verify quickly by making the velocity of the second 10...

9 years ago

P = W/t = mgh/t = 75*9.8*3/30 80% eff means divide above by 0.8

9 years ago

Your FBD should have weight down, friction up, and a normal force towards the center of the ride. The normal force is equal to the centripetal i.e. v^2/r. Mu (Fn) and as Bob showed this must be greater than mg. And v = (sqrt)(rg/mu). Important idea is tha...

9 years ago

Fd = mgh = 50*9.8*3

9 years ago

Energy doesn't care about direction. Just sum the two. kq3q1/r + kq3q2/r.

9 years ago

v = f (lambda) so lambda = v/f. f is two cycles per second (Hz).

9 years ago

Eq sin 30 = mg (make sure to change mass to kg). Find E. Now Eq cos 30 = T. A Free Body Diagram is most helpful in this problem.

9 years ago

Oscar Had A Headache Over Algebra. Sin x = Opp/ Hyp Cos x = Adj/ Hyp Tan = Opp/ Adj Opp = 2 sin24 Adj = 2 cos24

9 years ago

Google pythagorean theorem

9 years ago

Sum of Forces and Torques is zero. T1 + T2 = 650 + 200 200(2.1) + 650(number omitted) - T2(4.2) = 0 Use 2nd to find T2, plug into first for T1.

9 years ago

a) Pressure will be (rho)gh with rho being the density of water (which I don't know offhand) b)Mass flow rate (4500/seconds in a day) = (rho)Av. Where A is Pi r^2. c) I'm guessing this is a Bernoulli with (rho)gy1 = 1/2(rho)v^2 + (rho)gy2 Sorry this is a...

9 years ago

a)500 cos15 = T cos 25 b) T(vert) = 500 sin 15 + T(from part a) sin 25

9 years ago

KE = EPE 1/2m v^2 = 1/2k x^2 Make sure x is in meters.

9 years ago

Vs/Vp = Ns/Np Traditionally 1:10 would mean the secondary coil has 1/10 the number of turns.

9 years ago

it's a quadratic 9.8 t^2 + 8.95 t -29 = 0 Answer works out to 1.32 sec

9 years ago

Not enough info I believe. We need some info on the plates unless I missing an assumption.

9 years ago

Not having appendix F I'll ty to talk you through it. First you need the gravitational force between Earth and the sun. Get this from F = GmM/r^2. Next just use Coulomb's Law F (the one you just found) = kQ^2/r^2. In fact you can save a step and ignore th...

9 years ago

Physics? Think this is Chem. It's the point where the three phase transition lines meet I think.

9 years ago

Looks good

9 years ago

For circular opening sin(theta) = 1.22 (lambda)/D To get lambda use v = f (lambda) i.e. 343 = (lambda)/1800.

9 years ago

They seem to have told you the answer...Pythagorean. PS. Should have said they act at right angles.

9 years ago

[v sin40]^2 = 2 (9.8) 2.4. This assumes vy final is zero as it crosses the goal. v = sqrt(2*9.8*2.4)/sin40

9 years ago

See earlier 1.3 sec

9 years ago

Wow. I'm not even sure I can explain this one. Here goes. In x initial is vo cos30, in y, vo sin30. x velocity stays constant y velocity is given by final^2 - initial^2 = 2*9.8*42.9. Final TOTAL velocity is pythogorean sum of x final and y final. Set that...

9 years ago

Do kq1q2/r for each pair. Sum the three.

9 years ago

The two will have the same x velocities (19.2 cos29.2), and the distance traveled will be that number times t. Find the two times by solving in y: .5*-9.8t^2 + 19.2sin29.2t = -39.2 and .5*-9.8t^2 - 19.2sin29.2 = -39.2

9 years ago

v = vo + at. Plug. Chug.

9 years ago

See previous response...

9 years ago

No forces, no friction, no change in velocity. It will stay 22.5 forever.

9 years ago

This question is wrong. Mars' GRAVITY is about .38 Earth's. It's radius is about .54 Earth's. So by universal gravitation g = GM/r^2 g = G .11 / (.54 r)^2 = .11/.54^2 *9.8

9 years ago

x = .5(vo+v)t

9 years ago

2^2 - 1.3^2 = 2a(.03) (meters). Compare to 9.8. Use any other equation of motion to find t.

9 years ago

Ummmm. No idea.

9 years ago

Not sure I understand this. Yes energy will have to come from the water to raise the temp of the ice from -15 to 0, but it won't cause any phase change of the water , there's too much of it. .884 - .450 = .434 kg. Unless I'm really missing something...

9 years ago

y = .5 a t^2 so 1.5 = .5*9.8 t^2. Solve for t. b) x = vt so 3= v t (you got in part a)

9 years ago

Who's Felix? And where does he think he's going?

9 years ago

The electrons absorb quanta (packets) of energy. E = hf where h is Planck's constant.

9 years ago

A higher work function means the electron is less likely to be emitted. The actual KE of the emitted electron is determined by the frequency of the incoming light.

9 years ago

1) v = vo + at v = 40 - 9.8*3 2) x = v^2/(2*9.8) t = vo/t 3) t = twice the max height time.

9 years ago

1. Neither. Pushing down does nothing, pulling up moves it vertically. 2. Amazing Kreskin abilities fading... 3. The angle is 23.6o so Normal force is 10*9.8*cos23.6 and the friction force is .3 * Normal force. Friction force will give you work lost to fr...

9 years ago

Obvious reference to the great god "fissure". A Sumerian deity who could just about anything.

9 years ago

Could swear eV IS a unit of energy...

9 years ago

A railway? Don't know about a railway. If we're talking about using the normal force to keep circular motion we're talking about sin-1(v^2/(gr)) which is about 2.2 degrees. Then 1.6 sin2.2 will give the elevation.

9 years ago

F = kq1q2/r^2 = 9e9*1e-6*-1e-6/3^2. They exert the same force on each other (but opposite directions)

9 years ago

It depends on how high you pull it up. mgh = .5 m v^2 so vmax = sqrt(2*9.8*h) There's your general solution.

9 years ago

va cos55 + vb cos35 = 1.75*3.5 va sin55 = vb sin35 Solve the second eq for either variable, then plug it into the first eq. Voila.

9 years ago

Shells can be treated as point charges so this question simplifies considerably. E = kQ/r^2. To find Q multiply the charge density times the volume. Which by the way is a bit silly because all the charge is on the shell. Whatever. 2.22 * 4/3 pi r^3. You c...

9 years ago

q = kappa epsilon A V/d q = 5.3* 8.8e-12 * 4.5e-9 *62e-3 / 1e-8 Once you have this number the charge of a single ion (1.67e-19) to find out how many charges you need.

9 years ago

If you solve the two equations (x and y)of projectile motion to find range [in y direction t = 2 vo sin(theta)/g which substitutes into the x direction R = vo cos (theta) t, then use trig identity 2 sin(theta)cos(theta) = sin(2 theta)] you get R = v sin(2...

9 years ago

2T sin45 = 100*9.8

9 years ago

a = F/m = (41.8-24.4)/ 4.26 Get the mass by undoing mg: 41.8/9.8

9 years ago

I think you mean find the force between two electrons, right. You've given us the distances. F = kq1q2/r^2 q1 = q2 = 1.6e-19, k = 9e9. Now just substitute each distance in.

9 years ago

fo = fs (1+vo/v)/(1+vs/v) 40 = 40.4 (1+5/343)/(1+vs/343) I get about 8.5 so the bat dies of hunger.

9 years ago

A litter? Well you're going to need the weight of a litter of milk and you'll multiply that by the length of your forearm.

9 years ago

The two trucks meet in .625 hrs. At that time truck A has moved 25 km. The car has moved 37.5 km. So the two are 12.5 km apart.

9 years ago

For water Q = mc(Tf-Ti) (I don't know the specific heat of water offhand, get the mass using density). This must be equal to the heat required to melt the ice: Q = mL (don't know the latent heat of fusion for water either. Solve for mass of ice, use densi...

9 years ago

Three steps Q = mc(Tf-Ti) to get 16 to zero for both water and aluminum Q = mL to turn water to ice Q = mc(Tf-Ti) to get zero to -8 for both water and aluminum. Don't know specific heats of water, ice or aluminum offhand and I don't know latent heat of fu...

9 years ago

Volume of H2O = .03 * 2000 *3000 mass = density * vol Q = mL You'll need to look up latent heat of vaporization for water.

9 years ago

Initial velocity in the y direction must be. v^2= 2 (9.8) 3.7 or v = sqrt(2*9.8*3.7) Divide this by sin45 to get the total init vel

9 years ago

From the x direction: t = 56/40cos24 Use this t in the y equation: y = 40sin24 t - .5*9.8*t^2

9 years ago

1. Multiply both sides by 4 2. Divide both sides by m 3. Subtract k from both sides.

9 years ago

Let x = distance to town, t1 = time to town and t2 = return time. x = v1t1 = v2t2 so 36t1 = 48t2 Which seems like you can't solve it. But t2 = 3.5 - t1. So 36t1 = 48(3.5-t1). Solve for t1 and put back into x = 36(t1)

9 years ago

2.2* (5/3)

9 years ago

The outer electrons are further away in XeFl. They can be persuaded to react (although begrudgingly)

9 years ago

Careful. The Normal force is exerted by the table, not gravity.

9 years ago

Time to ditch quiz check on connexus. 2 is NOT D.

9 years ago