A certain spring stores 10.0J of potential energy when it isstretched by 2.00cm from its equilibrium position.How much potential
1 answer:
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Answer:
w=3.05 rad/s or 29.88rpm
Explanation:
k = coefficient of friction = 0.3900
R = radius of the cylinder = 2.7m
V = linear speed of rotation of the cylinder
w = angular speed = V/R or to rewrite V = w*R
N = normal force to cylinder
N=
These must be balanced (the net force on the people will be 0) so set them equal to each other.
There are 2*pi radians in 1 revolution so:
So you need about 30 RPM to keep people from falling out the bottom
a) 779 kg m/s
The momentum of an object is given by:
p = mv
where
m is the mass of the object
v is its velocity
For the fullback before the collision,
m = 95 kg
v = 8.2 m/s
Therefore, his momentum was:
b) -779 kg m/s
After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,
p' = 0
The initial momentum of the fullback was
p = 779 kg m/s
Therefore, his change in momentum is
where the negative sign indicates that the direction is opposite to the initial direction of motion.
c) -779 kg m/s
Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:
where
is the initial momentum of the fullback
is the initial momentum of the tackle
p' is the final combined momentum after the collision
We already know that
Therefore, we can find the tackle's original momentum:
where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.
e) -6.1 m/s
To find the velocity of the tackle, we can use again the equation of the momentum:
p = mv
where here we have
is the original momentum of the tackle
m = 128 kg is his mass
Solving the equation for v, we find the tackle's original velocity:
So, he was moving at 6.1 m/s in the direction opposite to the fullback.