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</p>
<p>This is a F=ma problem. You have F at 14.5 N, m = 22.5, and a = ? (in case you are confused as to how I got m, I simply divided the weight (which is equal to mg) by g to get m)</p>
<p>so F = ma</p>
<p>14.5 = 22.5a</p>
<p>14.5/22.5 = .644 m/s^2</p>
<p>Hope I helped.</p>
<p>thats wat I got 2 but the ans is 0.597</p>
<p>Could the difference have resulted from approximations/rounding? IMO that’s not a significant difference</p>
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</p>
<p>I just googled it. The reason the answer is .597 is because there is a 19 degree angle.</p>
<p>So force = cos (19) * 14.5 = 13.71</p>
<p>Now it becomes F = ma again.</p>
<p>13.71 = 22.9358a</p>
<p>a = 13.71/22.9358 = .5978 m/s^2</p>
<p>Hope I helped.</p>
<p>that sounds super right tho 
ty</p>
<p>I wanna make sure these statements are true:
When force is pushing @ an angle, normal force = (Mg + Fsin0)
When force is pulling @ an angle, normal force = (Mg – Fsin0)</p>
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</p>
<p>Try not to memorize things for this exam–it will only confuse you. There are very few instances in physics when you can assert that something will definitely happen.</p>
<p>First of all, the language you’re using is unclear. The force can most definitely push and be equal to Mg - Fsin0; similarly, it can pull and be equal to Mg + Fsin0. </p>
<p>Just think about a force diagram when determining what the normal force will be. If there is no force pulling upward or downward, the force normal = weight, right? So let’s change it up. If there is a force pulling upward (or pushing downward), then you know that the vertical component of the force must be subtracted from the weight (because in this case it would be weight = force normal + vertical component of pushing/pulling force). If there is a forcing pulling downward or pushing upward, the opposite is true–the vertical component of the force must be added to the weight (it would be weight + vertical component of force = force normal).</p>
<p>The free body is king. Draw one on every force problem. It’s complicated in words, but when you see it, it becomes so obvious that you’ll laugh at yourself for finding it difficult!</p>
<p>I hope I helped.</p>
<p>thx IRJunkie, that helps a lot.</p>
<p>A certain crane can provide a maximum lifting force of 25 000 N. It hoists a 2000-kg load starting at ground level by applying the maximum force for a 2-second interval; then, it applies just sufficient force to keep the load moving upward at constant speed. Approximately how long does it take to raise the load from ground level to a height of 30 m?</p>
<p>An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to complete. 31. What is the radius of the curve that the plane follows in making the turn?</p>
<p>anyone really having trouble with physics mc? what are some good helpful hints and tips?</p>
<p>
</p>
<p>Answer: Approximately 7 s</p>
<p>Solution:</p>
<p>Upward force = Lifting Force - Weight = 25000 N - (2000 kg) 10m/s/s= 5000 N</p>
<p>F= 5000 N
a= 5000 N/ 2000 kg = 2.5 m/s/s</p>
<p>Since the max force is applied for 2 seconds, the mass travels:</p>
<p>x= .5 at^2= 5 m
and
v= at= 5 m/s</p>
<p>x = x0 + vt + 1/2at^2
25 m= 5t
t=5</p>
<p>Thus, total t= 5 + 2=7</p>
<p>Anyone care to explain something about thermo? like when they ask you if heat should be added, removed, or neither, or what is the sign of the heat/work done by/to the gas? thanks</p>
<p>
</p>
<p>Equation is U = Q + W, internal energy = heat added + work done on it; so if the gas does work, W would be negative, etc.</p>
<p>A rock is whirled on the end of a string in a horizontal circle of radius R with a constant period T. If the radius of the circle is reduced to R/2, while the period remains T, what happens to the centripetal acceleration of the rock?</p>
<p>Dude, you’re shooting out questions before previous ones are answered…</p>
<p>Question:An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to complete. 31. What is the radius of the curve that the plane follows in making the turn?</p>
<p>Answer: 1098 m</p>
<p>Solution:
x=vt= 1725 m
pi/2 x r= 1725
r=1098 m</p>
<p>Well, it’s asking for centripetal acceleration, which is defined as a = (v^2)/r. So we know that the radius is now R/2; what’s the velocity, then?</p>
<p>So now the circle has radius R/2, with period T, so the rock travels a distance of piR [2piR/2] meters/T. That’s the velocity; just put that into (v^2)/(R/2) and that should be it [the new acceleration]. If you have to find the difference between the two, that’s relatively simple using similar methods.</p>
<p>Question: A rock is whirled on the end of a string in a horizontal circle of radius R with a constant period T. If the radius of the circle is reduced to R/2, while the period remains T, what happens to the centripetal acceleration of the rock?</p>
<p>Answer: The centripetal acceleration decreases by a factor of 2.</p>
<p>A little off topic but it says official thread…does anyone else have PR and find the MC to be ridiculously hard? I think it may be more difficult, my teacher gave us a prac test in school which was provided by collegeboard and I got like a 4 but on the princeton one i think i must’ve gotten at least 80% wrong</p>
<p>PR tends to make things seem harder than they actually are, so to answer your question, yes.</p>