The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf
1 answer:
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1) 29.4 N
The force of gravity between two objects is given by:
where
G is the gravitational constant
M and m are the masses of the two objects
r is the separation between the centres of mass of the two objects
In this problem, we have
(mass of the Earth)
(mass of the box)
(Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)
Substituting into the equation, we find F:
2)
Let's now calculate the ratio F/m. We have:
F = 29.4 N
m = 3.0 kg
Subsituting, we find
This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol .
We can prove that this is the acceleration of the object: in fact, according to Newton's second law,
where a is the acceleration of the object. Re-arranging,
which is exactly equal to the quantity we have calculated above.
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} }