Sorry about that, but wouldn't it be N/s? Also, what unit is momentum in? I had a question on a test the other day and was uncertain... as you can tell, impulse and momentum are concepts I'm not as familiar with.
No, it's N*s. As you said, the unit of momentum is kg*m/s. Since Impulse is the change in momentum, it will have the same unit. A Newton is kg*m/s^2. Therefore, the unit for momentum is the Newton times a unit of time.
Oooohh, I didn't realize that a Newton is how much a kilo can be accelerated! Thanks for the tip!
Re: Hovercraft B/C
Posted: March 5th, 2017, 11:05 am
by HandsFreeCookieDunk
Zioly wrote:
HandsFreeCookieDunk wrote:
Zioly wrote:
Sorry about that, but wouldn't it be N/s? Also, what unit is momentum in? I had a question on a test the other day and was uncertain... as you can tell, impulse and momentum are concepts I'm not as familiar with.
No, it's N*s. As you said, the unit of momentum is kg*m/s. Since Impulse is the change in momentum, it will have the same unit. A Newton is kg*m/s^2. Therefore, the unit for momentum is the Newton times a unit of time.
Oooohh, I didn't realize that a Newton is how much a kilo can be accelerated! Thanks for the tip!
Yup, right from F=ma
Re: Hovercraft B/C
Posted: March 10th, 2017, 4:04 pm
by Zioly
Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.
You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.
Re: Hovercraft B/C
Posted: March 13th, 2017, 7:42 pm
by RestingDoll
Zioly wrote:Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.
You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.
v^2=vi^2+2a(delta_y)
v^2=0-2g(26)
v^2=510.12
v=22.5858 m --> v~23 m
Re: Hovercraft B/C
Posted: March 13th, 2017, 7:44 pm
by RestingDoll
Zioly wrote:Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.
You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.
Wait lol how do I hide it. Edit: nvm, figured it out
Re: Hovercraft B/C
Posted: March 13th, 2017, 8:24 pm
by Zioly
RestingDoll wrote:
Zioly wrote:Sorry, I thought that since I got it wrong, someone else could have a go at the right answer and put up the next question.
You drop a ball from a 26 meter tall building on earth. Find the velocity when the ball hit the ground. Assume no air resistance and gravity as 9.81. Show all work.
v^2=vi^2+2a(delta_y)
v^2=0-2g(26)
v^2=510.12
v=22.5858 m --> v~23 m
All correct, except for that one little misclick. (You typed "-" instead of "+" in line 2. Your go!)
Re: Hovercraft B/C
Posted: March 14th, 2017, 12:10 pm
by RestingDoll
A small 60 g point mass starts sliding down a frictionless surface at 30 cm in height, then sticks to the end of a uniform rod of mass 120 g and length 40 cm. The rod rotates about a point "O" lengthwise and its moment of inertia about that point is given by 1/3 mr^2. After the point mass sticks to the rod, it pivots to an angle theta, before momentarily stopping. Find theta. (Hint: Don't forget that the potential energy stored after it pivots is based on the center of mass of the entire system.)
Re: Hovercraft B/C
Posted: April 18th, 2017, 8:43 am
by Tom_MS
RestingDoll wrote:A small 60 g point mass starts sliding down a frictionless surface at 30 cm in height, then sticks to the end of a uniform rod of mass 120 g and length 40 cm. The rod rotates about a point "O" lengthwise and its moment of inertia about that point is given by 1/3 mr^2. After the point mass sticks to the rod, it pivots to an angle theta, before momentarily stopping. Find theta. (Hint: Don't forget that the potential energy stored after it pivots is based on the center of mass of the entire system.)
75.5 degrees. If I'm correct, the 1/3mr^2 only matters to determine where the center of mass is.
Re: Hovercraft B/C
Posted: April 21st, 2017, 3:34 pm
by RestingDoll
Tom_MS wrote:
RestingDoll wrote:A small 60 g point mass starts sliding down a frictionless surface at 30 cm in height, then sticks to the end of a uniform rod of mass 120 g and length 40 cm. The rod rotates about a point "O" lengthwise and its moment of inertia about that point is given by 1/3 mr^2. After the point mass sticks to the rod, it pivots to an angle theta, before momentarily stopping. Find theta. (Hint: Don't forget that the potential energy stored after it pivots is based on the center of mass of the entire system.)
75.5 degrees. If I'm correct, the 1/3mr^2 only matters to determine where the center of mass is.
Interesting... I think you may be forgetting that the center of mass of the system is different than the center of mass of the rod.
