Finding tension -- really basic
An automobile engine has a weight whose magnitude is 3150 N. The engine is positioned above an engine, and is positioned by a rope.
There is an illustration given, but it's essentially just saying that the angle concerning T1 is 10 degrees...
A student is skateboarding down a ramp that is 6.2 m long and inclined at 13° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 2.1 m/s. Neglect friction and find the speed at the bottom of the ramp.
I got the right answer! Thank you so much for your help and patience :)
I just have one last question, since I never learned this center of mass stuff: can I use that in any force problem when two masses are given and both objects are accelerating (towards each other)?
So, Dt=460 m(6100 kg)/(6100+3700)
Which = 286.3
And then Da= 460 m(3700 kg)/(6100+3700)
Which = 173.7
Which both add up (when I don't round) to be 460.
So, now that I have those, I only really need to use either Dt or Da, with the associated acceleration to find time, right?
Wait, so would this work?
I make the tug my centre of mass, so that the asteroid has to travel 460 m (as stated in the problem). Then I find it's acceleration (the same one as in my first post) and find how long it takes, using that acceleration, to travel 460 m? Because that's what I did the...
I've never come across this =/ I don't know if it was in the lesson and I missed it or if it wasn't covered. I never took physics in high school, either.
I don't think I quite understand the first two paragraphs here (well, how to apply them). But am I right in restating what you say as that I...
I missed the class in which this was discussed and these "reference frames" aren't really mentioned in my textbook... Can I ask you to elaborate on that?
I found the acceleration for both the asteroid and the tug, but I wasn't sure what to do when I had both of them.
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3700-kg space tug and a 6100-kg asteroid. Using their ship's engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 460 m...
Oops, sorry. I should have clarified that I did that. That's how I got the resultant vector; I found the components of the vector that had a force of .20 N. The vector in the x direction was .12 N and the vector in the y direction was -.16 N.
A duck has a mass of 2.7 kg. As the duck paddles, a force of 0.13 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.20 N in a direction of 54° south of east. When these forces begin to act, the velocity of the duck is 0.13 m/s in...
Alright, so using that I found that at 20 degrees, vx = 21 m/s
So, I need to find v, or vy.
to do that, I plugged in 21 = v cos 20
So I found that v = 22 m/s
Then I plugged that into 222 = 212 + vy2
So I got vy = 7.6. So I plugged into vy(final) = vy(initial) + ayt
And I got t = .13...
Ah, thanks! I was wondering how to enter those!
See, that's where I'm confused. Acceleration would be constant in the y direction (-9.80 m/s2), and then acceleration would be 0 in the x direction... but, then, nothing else is constant.
In the javelin throw at a track-and-field event, the javelin is launched at a speed of 26 m/s at an angle of 35° above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required...