Recent content by PureEnergy

As you figured out in part a, the smallest force should only require the boy to balance out the y component. In part b, I believe the question is asking for the angle this force makes with respect to the x-axis and since this force is downward, the angle should be 270. For part c, you just need...

I believe my method is correct and see no reason for integration (assuming you already have the derived basic kinematic equations). I suspect one of your given values is incorrect. It's possible the teacher may have made a mistake for one of the numbers.

The motorist must stop within a time of 38/18, since this is how long it will take her to cover 38 meters. The minimum time it will take her to stop occurs when she deaccelerates at a constant -4.50 until the car stops. The maximum reaction time will be the difference of these two times. Using...

Both your equations are wrong. When you integrated the acceleration, you forgot about the constant of integration or the initial velocity.

To find the maximum height, you should only be concerned with motion in the y-direction. Physically, you want to think of how long will it take for gravity to slow down the ball until it stops. Once you find this time, you can plug it in the y analog the equation you have in part a to find the...

cos^{2}(4x)*cos^{10}(4x) * sin(4x) dx = cos^{12}(4x)sin(4x)dx I believe the mistake you are making is multiplying cos^{2}(4x) once with the cos^{10}(4x) and another time with sin(4x) dx . This is incorrect because the two latter terms are multipled together.

Since the ball is thrown downward it has a velocity of -6.35 and not 6.35, be sure to always check the direction. For the last part, don't forget that you begin with a velocity of -6.35 m/s.

Technically, number 8 should be Fred Cooper since it's his processed voice Hawking uses when he speaks.

Along with Doc's equation, you'll want to use [itex]\Delta U = Q + W[/tex]

This problem can be solved using \Delta S = nC_Pln\frac{V_2}{V_1}+nC_Vln\frac{P_2}{P_1} All you need to do now, is to recall the relationship between an ideal gas and the heat capacity. You don't need the temperature to solve this problem.

Doc's right, the third step should be negative so the net work is around -550.

daisyi - if you add up all your work terms again, you'll see it will not give you -365. The calculation for all the individual work terms are correct, except for some rounding errors. When I did the calcuations I get: 1201.18 - 1501.47 + 249.42 = -50.87 or -50 after rounding.

We could convert the m by saying it is equivalent to moles*(molecular weight) and since n =1, then m = MW. It shouldn't be a problem if your answer has MW and R because they are both constants.

For a monotonic ideal gas Cv = 3/2R. Use the the ideal gas equation to change from temperature to pressure and volume. Since in part b you have an isochoric process, the volume will be fixed and only the pressure will change. This will give you the following equation \Delta U = 1.5m\Delta{p}V...

To solve the problem you can use this equation to find the molar entropy \Delta S = \int_{T_1}^{T_2} C_v \frac{dT}{T} Multiply this result by the numer of moles (1.2/12) to get entropy in terms of energy/time.