Cells that remain as sources of other cells are called
1 answer:
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1) The most potential energy is at position A
2) The most kinetic energy is at position C
Explanation:
1)
The gravitational potential energy is the energy possessed by an object due to its position in a gravitational field, and it is given by the equation
where
m is the mass of the object
g is the acceleration of gravity
h is the height of the object relative to the ground
From the equation, we see that the potential energy is directly proportional to the heigth of the object: therefore, the roller coaster in this problem will have the most potential energy at its highest postion, so at position A.
2)
The total mechanical energy of the roller coaster at any point along the track is given by
where
PE is the potential energy
KE is the kinetic energy
Assuming there is no friction, the mechanical energy E is constant. This means that when PE increases, KE decreases, and when PE increases, KE decreases.
Therefore, the cart will have maximum kinetic energy when the potential energy is at minimum: and since the potential energy is directly proportional to the height of the track, this will occur at the lowest position, so at position C.
Learn more about kinetic and potential energy:
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Answer:
x = A cos (w \sqrt{2y_{o}/g})
a) maximun Ф= \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) minimun Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Explanation:
For this exercise let's use kinematics to find the time it takes for the mass to reach the floor
y = y₀ + v₀ t - ½ g t²
as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)
0 = y₀ - ½ g t²
t =
The bucket-spring system has a simple harmonic motion, which is described by
x = A cos wt
in this expression we assumed that the phase constant (Ф) is zero
let's replace the time
x = A cos (w \sqrt{2y_{o}/g})
this is the distance where the system must be for the mass to fall into it.
a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes
w =
In the initial state
w = \sqrt{k/2}
When the mass changes
w ’= \sqrt{k/3}
the displacement in each case is
x = A cos (wt)
for the new case
x ’= A cos (w’t + Ф)
the phase constant is included to take into account possible changes due to the collision of the mass.
we see that this maximum expressions when the cosine is maximum
cos (w´t + Ф) = 1
w’t + Ф = 0
Ф = -w ’t
Ф = -
\sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) the function is minimun if
cos (w’t + fi) = 0
w’t + Ф = π / 2
Ф = π / 2 - w ’t
Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }