Ok. I got it. I came up with this formula that substitutes the inelastic equation because like you said, there is more than once concept. Here is what I used:
Kf = .5(m2) * (2m1 / m1 +m2)^2 * vo^2
m2 is the weight of the nail in kg
m1 is the the weight of the hammer in kg
I got K= .3933 J...
Hi! Here's the question: I really need a starting block for this because I really don't know where to start.
The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by a 15 g nail when it is struck by a 550 g hammer moving...
ok...so, how would I figure (b) out? Since the amplitude is reached 4 times in a second, and the time interval is 3 seconds, and the amplitude is 12 cm, how would you figure out the distance traveled?
A 4.0 Hz wave with an amplitude of 12 cm and a wavelength of 30.0 cm travels along a stretched string.
(a) How far does the wave travel in a time interval of 3.0 s?
m
(b) How far does a knot on the string travel in the same time interval?
m
I got the answer to part (a) by using...
After getting a drink of water, a hamster jumps onto an exercise wheel for a run. A few seconds later the hamster is running in place with a speed of 0.8 m/s. Find the work done by the hamster to get the exercise wheel moving, assuming it is a hoop of radius 0.10 m and mass 5.9 g.
I'm not...
Hi everyone. I have another problem that is just stumping me. I posted this earlier and still can't understand it. Here it is...
A 1-kg ball starting at h = 6.1 meters slides down a smooth surface where it encounters a rough surface and is brought to rest at B, a distance 18.3 meters away. To...
A 2.0 kg block is falling at a speed of 11 m/s and is 18 meters above the spring. The spring constant is 4000 N/m, to the nearest hundredth of a meter what height will the block rise after it hits and leaves the spring?
Help please! I can find the kinetic energy of the falling block by...
The surface is a slope that evens out at the end...like a waterslide...by a smooth surface, I believe that means "frictionless"...the 15.8 meters represents the distance when you hit the rough surface to when you stop. The rough surface starts at the "bottom of the slide" and runs along the...
Question:
A 1-kg ball starting at h = 8.6 meters slides down a smooth surface where it encounters a rough surface and is brought to rest at B, a distance 15.8 meters away. To the nearest joule what is the work done by friction?
Can you pls. help with this question? Do you need to find...
Nevermind...(a) would equal -9.81 because its the acceleration due to gravity. I solved for x, but then I noticed I had to add on the 30 m. of the cliff. I came up with 32 m. Thanks a lot for the help!
For part (a) I used Ui + Ki = Uf + Kf ...which gave me
mgy1 + .5mvi^2 = mgyf +.5mvf^2
I Then solved for vi
For part (b) I'm not sure what equation to use. I was thinking to use this one:
v^2 = vo^2 + 2a(delta)x
the final velocity should equal 0 right? because the rock is on...
A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 30 m high. When it hits the ground at the base of the cliff the rock has a speed of 25 m/s.
(a) Assuming that air resistance can be ignored, find the initial speed of the rock.
(b) Find the greatest height of the...
I'm still a little confused. What value do you have to find first? If you use Spring-Potential Energy equation, how would you find the speed of the arrow?
No, not really. If you found the Force of the "draw-back" bow, could you figure out the speed of the arrow? Because it says when the arrow is released, does that mean you don't need to take into account the weight of the arrow?
Help pls! Here is the question:
"Assuming you have a bow that behaves like a spring with a spring constant of 80 N/m and you pull it to a draw of 51 cm, to the nearest tenth of a m/s, what is the speed of the 81-gram arrow when it is released?"
What equations do you use? I'be been trying...