What is the force acting on a 10kg object that accelerates from 5 m/s to 20 m/s in 5s?
1 answer:
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Answer:
The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.
Explanation:
Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively
and the mass of big rock be 'M'
Initial momentum of the system equals
Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'
Thus the final momentum of the system is
Equating initial and the final momenta we get
Now since the surface is frictionless thus the energy is also conserved thus
Similarly the final energy becomes
\
Equating initial and final energies we get
Solving i and ii we get
Using this in equation i we get
Thus putting v = -v' in equation i we get V' = 0
This implies Smaller stone rebounds while as larger stone remains stationary.
Answer:
Explanation:
Use the one-dimensional equation
Δx = where delta x is the displacement of the object, v0 is the velocity of the object, a is the pull of gravity, and t is the time in seconds. That's our unknown.
Δx = -2 (negative because where it ends up is lower than the point at which it started),
, and
a = -9.8
Filling in:
and simplified a bit:
this should look hauntingly familiar (a quadratic, which is parabolic motion...very important in physics!!). We begin by getting everything on one side of the equals sign and solving for t by factoring:
(the 0 is also indicative of the object landing on the ground! Isn't this a beautiful thing, how it all just works so perfectly together?)
When you factor this however your math/physics teacher has you factoring you will get that
t = 1.3 sec and t = -.31 sec
Since we all know that time can NEVER be negative, it takes the ball 1.3 sec to hit the ground from a height of 2 m if it is rolling off the shelf at 5 m/s.