In order to decrease the friction on the slide,
we could try some of these:
-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide. The oil oozes
down the slide and keeps the whole slide greased.
-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.
-- Spray the whole slide with soapy sudsy water, every 30 minutes.
-- Drill a million holes in the slide,and pump high-pressure air
through the holes. Make the slide like an air hockey table.
-- Keep the slide very cold, and keep spraying it with a fine mist
of water. The water freezes, and a thin coating of ice stays on
the slide.
-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide. If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.
-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide. Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.
Answer:
Fluid fricton goes to Static friction and sliding friction goes to rolling friction
Explanation:
Answer of your question is in this photo
Answer:
option C
Explanation:
given,
Force on the object = 10 N
distance of push = 5 m
Work done = ?
we know,
work done is equal to Force into displacement.
W = F . s
W = 10 x 5
W = 50 J
Work done by the object when 10 N force is applied is equal to 50 J
Hence, the correct answer is option C
Answer:
2 m/s and -2 m/s
Explanation:
The object travels with an angle of
60.0°
with the positive direction of the y-axis: this means that it lies either in the 1st quadrant (positive x) or in the 2nd quadrant (negative x).
If it lies in the 1st quadrant, the value of vx (component of v along x direction) is:
![v_x = v cos \theta = (4.0 m/s) cos 60.0^{\circ}=2 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20v%20cos%20%5Ctheta%20%3D%20%284.0%20m%2Fs%29%20cos%2060.0%5E%7B%5Ccirc%7D%3D2%20m%2Fs)
If it lies in the 2nd quadrant, the value of vx (component of v along x direction) is:
![v_x = -v cos \theta = -(4.0 m/s) cos 60.0^{\circ}=-2 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20-v%20cos%20%5Ctheta%20%3D%20-%284.0%20m%2Fs%29%20cos%2060.0%5E%7B%5Ccirc%7D%3D-2%20m%2Fs)