Sounds like an interesting ideamattruff wrote:Are we allowed do put Lego wheels on the sides of our hovercraft to help the hovercraft roll against the sides ?
. Not too sure if it’d be allowed but try asking on the actual hover forum I’m sure some discussion will spur on it.
A hammer with mass m is dropped from rest from a height h above the earth's surface. This height is not necessarily small compared with the radius Re of the earth. If you ignore air resistance, derive an expression for the speed v of the hammer when it reaches the surface of the earth. Your expression should involve h, Re, and me, the mass of the earth.Riptide wrote:Nicely done! Your turnJustin72835 wrote:Riptide wrote:A circle rotates clockwise at 100 rpm and has a constant counterclockwise acceleration of 4 rad/s^2 for 5.5 seconds. What is the angular displacement?100 rpm = 10.47 rad/s (multiply by pi/30). Then using rotational kinematics (use the displacement equation), you get an answer of 2.9 rad counterclockwise.
Justin72835 wrote:A hammer with mass m is dropped from rest from a height h above the earth's surface. This height is not necessarily small compared with the radius Re of the earth. If you ignore air resistance, derive an expression for the speed v of the hammer when it reaches the surface of the earth. Your expression should involve h, Re, and me, the mass of the earth.Riptide wrote:Nicely done! Your turnJustin72835 wrote:100 rpm = 10.47 rad/s (multiply by pi/30). Then using rotational kinematics (use the displacement equation), you get an answer of 2.9 rad counterclockwise.
When the hammer moves towards the Earth, its gravitational potential energy becomes a bigger negative number. This results in the "release" of some kinetic energy in the form of the hammer moving by the Conservation of Energy.
[math]v = \sqrt{\frac{2GM_{e}}{R_{e}} - \frac{2GM_{e}}{R_{e} + h}}[/math] Good job; I guess it's your turn now!Adi1008 wrote:When the hammer moves towards the Earth, its gravitational potential energy becomes a bigger negative number. This results in the "release" of some kinetic energy in the form of the hammer moving by the Conservation of Energy. [math]v = \sqrt{\frac{2GM_{e}}{R_{e}} - \frac{2GM_{e}}{R_{e} + h}}[/math]
A uniform stick of lengthJustin72835 wrote:Good job; I guess it's your turn now!Adi1008 wrote:When the hammer moves towards the Earth, its gravitational potential energy becomes a bigger negative number. This results in the "release" of some kinetic energy in the form of the hammer moving by the Conservation of Energy. [math]v = \sqrt{\frac{2GM_{e}}{R_{e}} - \frac{2GM_{e}}{R_{e} + h}}[/math]
Adi1008 wrote: A uniform stick of lengthis fixed at one end and is free to rotate in the plane of the computer screen. It is initially released at an angle
to the vertical. When the stick has fallen to the point where it hangs straight down, the speed at the tip is
. What is
Drawing a diagram makes it easier to answer this question. The initial height is (L/2)*(1+cos[math]\theta[/math]). Using conservation of energy we get 3L(1+cos[math]\theta[/math]) = v^2, which results in [math]\theta[/math] = arccos(v^2/(3Lg)-1).
Looks good bro. Your turn!Riptide wrote:Adi1008 wrote: A uniform stick of lengthDrawing a diagram makes it easier to answer this question. The initial height is (L/2)*(1+cos[math]\theta[/math]). Using conservation of energy we get 3L(1+cos[math]\theta[/math]) = v^2, which results in [math]\theta[/math] = arccos(v^2/(3Lg)-1).
Riptide wrote:A 8 gram bullet is fired horizontally with a speed of 400 m/s into a 1.2 kg block of wood at rest on the floor. If the coefficient of friction between the block and the table is 0.2, A) how far will the block slide and B) what fraction of the bullet's energy is dissipated in the collision itself?
[math]1.79m[/math], [math]0.9933[/math]
Nice work your turn!Justin72835 wrote:Riptide wrote:A 8 gram bullet is fired horizontally with a speed of 400 m/s into a 1.2 kg block of wood at rest on the floor. If the coefficient of friction between the block and the table is 0.2, A) how far will the block slide and B) what fraction of the bullet's energy is dissipated in the collision itself?[math]1.79m[/math], [math]0.9933[/math]
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