It doesn't matter. If the slides are truly frictionless, then
your kinetic energy at the bottom will be equal to the
potential energy you had at the top, no matter what kind
of route you took getting down.
___________________________
The only way I can think of that it would make a difference
would be if the shallow slide were REALLY REALLY long,
and you didn't have anything to eat all the way down.
Then you might lose some weight while you're on the slide,
and your mass might be less at the bottom than it was at the
top. Then, in order to have the same kinetic energy at the
bottom, you'd need to be going a little bit faster.
But if it takes less than, say, two or three days, to go down the
long, shallow slide, then this effect would probably be too small
to make any difference.
-1.1m/s
Explanation:
Work out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V). If you have U, A and T, use V = U + AT. If you have S, U and T, use V = 2(S/T) - U.
49 J is the total kinetic energy. If a bowling ball of mass 7.3 kg and radius 9.6 cm rolls without slipping down a lane at 3.1 m/s. Kinetic energy is the energy an bowling ball has because of its motion.
Given: m = 7.3 Kg ; r = 9.4 cm = 0.094 m ; v = 3.1 m
Now total kinetic energy in this case is given by KE = Kinetic energy due to rotation + Kinetic energy due to translation
i,e KE = 1/2*m*v2 + 1/2*I*ω2 where I is the moment of inertia of the bowling ball about it's center and ω is the angular velocity
Now for pure rotation (without slipping) v = rω
also for the ball (solid sphere) I = 2/5*m*r2
Hence our kinetic energy becomes
KE = 1/2*m*v2 + 1/5*m*v2 = 7/10*m*v2
so KE = 0.7*7.3*(3.1)2 = 49.10 J = 49 J
Learn more about kinetic energy here
brainly.com/question/12669551
#SPJ4
Answer:
0.00001581 light years
Explanation:
distance divided by speed of light= 150,000,000/300,000= 500 seconds 8.3333
8.3 minutes
now divide this by 60 x 24 x 365.25. so you will get answer=
.000015823 light years.
26 neutralization reaction;
27 positive ions;
28 titrations;
29 bases;
30 buffers
