Answer:
Change in momentum,
Explanation:
It is given that,
Mass of the basketball, m = 601 g = 0.601 kg
The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s
It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :
So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.
Hi there!
(a)
Recall that:
W = Work (J)
F = Force (N)
d = Displacement (m)
Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.
To the nearest multiple of ten:
(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.
Thus:
(c)
Similarly, the normal force is perpendicular to the displacement, so:
(d)
Recall that the force of kinetic friction is given by:
Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
In multiples of ten:
(e)
Simply add up the above values of work to find the net work.
Nearest multiple of ten:
(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
Nearest multiple of ten:
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 / - 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.