Recent content by HumorMe81

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    Kinematics motion, projectile. This was an mcat problem. Thanks

    Oh, so the time is 5s then. 2s after the explosion and 3s to come back down? If the total time is 5s then horizontal distance is 15m/s*5s = 75m and the total distance between the two fragments after they hit the ground is 33m+75m = 108m. That's the answer listed in the book.
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    Kinematics motion, projectile. This was an mcat problem. Thanks

    Ok, so i figured out what I was doing wrong. The fragment moving left travels a vertical distance of 25m. Right before it starts coming back down, its velocity at the highest point would be 0. I used this info to calculate time on the downward flight: Y = Vi*t + 1/2gt^2 25 = 0 +...
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    Kinematics motion, projectile. This was an mcat problem. Thanks

    A firecracker with mass 100g is propelled vertically with a launch force of 1.2N applied over 5s, after which it explodes into two 50g fragments. One fragment goes horizontally at 15m/s and the other goes to the right at angle 53 degrees with the horizontal at speed 25 m/s. find the distance...
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    Motion in a vertical circle

    Ok, thanks a lot for your help!
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    Motion in a vertical circle

    A 0.8 kg ball is whirled on a string r = 0.4 meters long in a vertical circular path. At the bottom of the circle, the ball is h = 0.45 meters from the ground. At the top of the circle, the ball has a speed of 3 m/s. Assume that the total energy of the ball is kept constant. a. Calculate...
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    Friction on an inclined plane/pulley

    if you resolved this question, can u plz post the answer. i want to see if i solved it correctly. thanks.
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    Friction on an inclined plane/pulley

    the system could be moving to the right and decelerating.
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    Elevator question

    ok, if the acceleration were constant and the initial velocity were given to be 0, then i see some sense in solving this problem. And even then the answer wouldn't match any of the given answers.
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    Elevator question

    ok, so does that mean avg speed is = 2X/t? if so, then the answer is a. 2X + 1000 I still don't see the relationship between avg speed and avg acceleration.
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    Elevator question

    avg velocity comes to mind since the question says the elevator travels a distance x over a period of 10s. If i solve the problem using this information, the answer comes out to be 2X + 1000. But avg accleration is equal to delta v/ delta t, not avg velocity. And also, using avg velocity would...
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    Elevator question

    100 kg elevator acclerates up at rate avg a*. What's the avg force acting on the elevator if it covers a distance X over a period of 10s? a* denotes average accleration a. 2X + 1000 b. 100(a* + 1) c. 2a* + 1000 d. 2a*X I set up a free body diagram to show all the forces acting on the...
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    Standing Waves in a closed pipe

    I tried your recommendation in a problem in my mcat book and it works great. For some reason, the book doesn't discuss how to figure out nodes and antinodes. Thank you so much, really appreciate it!!
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    Standing Waves in a closed pipe

    So confused about standing waves in a closed pipe, which is open at one end and closed at the other. The closed end has a node while the open end has an antinode. To figure the wavelength, i use the formula: Lambda = 4L/n where n is the number of harmonic and can only be odd integers...
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    Work and Energy

    Ok, got it! Thank you so much! W'(work done by friction) = delta E = Fk*d
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    Work and Energy

    I used deltas to define the initial and final energies of the entire syste. Ei = Ui + KEi = 1/2kx^2 + 0 So, the initial energy = 1/2kx^2 only b/c the block has not been released yet. Finally, Ef = (Uf + KEf) - work done by friction This is how it makes sense to me. my mcat book...
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