I understand why you may think that, but no. Keep in mind your inertial frame of reference - although I'm not sure that really explains it very well...
After the first person jumps (we'll say to the right, which we'll call positive), the car begins moving to the left - which means the car's...
Part a is correct.
However, in part b, you have your final velocities (after the first person jumps) incorrect. Since, according to your equations, the car is pushed in the negative direction, the next person's v2 would be 2 m/s minus the car's velocity - not plus.
No, that is the correct answer. 11.38 m/s to be "precise," but you're only given two significant figures in your initial conditions. I've done it three ways, and it comes to 11.38 m/s each time. And 40 m/s is correct for a. Just now, I did it a 4th way, and I got 10.96 m/s - which still...
You converted your answer to slugs, not lbm. Personally, I loathe the idea of lbm. In practical use - although not entirely correct - lbf is equivalent to lbm. So, your original numerical answer of 1014 is correct - just different due to rounding in between steps.
The problem asked for the speed of the rock - you gave its velocity. Remember - speed is a scalar (independent of direction), and velocity is a vector (depends upon direction).edit: gneill and I were posting at the same time, apparently.
For an isothermal process with an ideal gas, the heat added (or subtracted - depending upon exactly what happens to the gas) is equivalent to the work done on (or by) the gas.
No. Since it's a double-shear pin - the force is equally spread across two shear points. Thus, the force at either individual point is 75000 N. Use that force, the shear strength, and the equation for area of a circle (since it's asking for diameter, we assume the pin's cross-section is a...
According to the way the question is worded - namely what do you know must be false - we only know that the velocity cannot be zero in the z-direction. As the question doesn't mention any acceleration, what do we know about any forces?
Two things: Since it's double-shear, you divide the shearing force by two, not multiply. Also, you seem to have gotten the equation for area of a circle incorrect - it's not d2. It's (pi*d2)/4.
The fact that it's a cantilever doesn't affect anything, really. It just means that the support isn't on the end. You still go about your static equilibrium equations the same way and work from there. If it helps, just think of it as an unknown force. Were this a fixed support (with only one...
What's the definition of exothermic?
(Hint: How does exo- differ from endo-thermic?)Also, since you're doing this in a lab - aren't you measuring the temperature of the solution?