A tennis ball bouncing on a hard surface compresses and then rebounds. The details of the rebound are specified in tennis regula
1 answer:
The velocity of the ball is 7 m/s
Explanation:
The motion of the ball is a free fall motion, so it means that the ball falls down under the effect of the force of gravity only. Therefore, it has a constant acceleration (acceleration of gravity, g), and we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the ball in this problem, we have:
u = 0 (initial velocity, the ball is dropped from rest)
(acceleration of gravity)
s = 2.5 m (vertical displacement)
Solving for v, we find the velocity at which the ball hits the concrete surface:
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Answer:
The potential for r > rb is equal to zero.
Explanation:
For r > rb, the potential is:
Then, the net potential is:
Answer:
the Zeeman effect without spin is three lines
Explanation:
The Zeeman effect is the result of the interaction of the magnetic field with the orbital angular momentum of the electrons, if we do not take the spin into account it is called the Normal Zeeman effect.
Therefore we only take into account the orbital moments (m_l) of the transition, from the selection rules of the refreshed harmonics, only the transition with
= 0, ± 1
ΔE
Let's analyze for the case of the Hydrogen atom
For a transition between levels n = 1 and n = 2 the values of m_l are n_f = 1 m_l = 0 and for n₀ = 2 m_l = 0, 1
so we only have two lines.
For transition n_f = 2 and n₀ = 3
n_f = 2 m_l = 0, 1
n₀ = 3 m_l = -1, 0, 1
There are only two lines plus the central line, so there are three spectral lines
for n_f = 3 and n_o = 4
n_f = 3 ml = -1, 0, 1
n₀= 4 ml = -2, -1, 0, 1, 2
Only transitions with Δm_l = ±1 are allowed, so there are only two transitions plus the central transition (Δm_l = 0), so there are only 3 spectral lines.
In summary, due to the selection rule of spherical harmonics, the greatest number of lines in the Zeeman effect without spin is three lines.
a) The work done by the tree is
b) The amount of force applied is 2880 N
Explanation:
a)
According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:
where
W is the work done on the car
m is the mass of the car
u is its initial speed
v is its final speed
For the car in this problem, we have:
m = 2000 kg
u = 12.0 m/s
v = 0 (since the car comes to a stop, after the crash)
Therefore, the work done by the tree on the car is:
The work is negative because it is done in the direction opposite to the direction of motion of the car.
b)
The work done by the tree on the car can also be rewritten as
where
F is the force applied on the car
d is the displacement of the car during the collision
In this situation, we have:
is the work done
is the displacement of the car during the collision
Solving the equation for F, we find the force exerted by the tree on the car:
Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.
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