Answer:
V = 6.23 cm³
Explanation:
given,
Length of the cylinder, h = 5 cm
Diameter of the cylinder, d = 1.26 cm
r = 0.63 cm
Volume of the cylinder = ?
We know,
V = 6.23 cm³
Volume of the cylinder is equal to V = 6.23 cm³
Given:
m = 0.32 kg
v = 11.5 m/s
To find:
1) Kinetic energy = ?
2) Work needed to stop the ball = ?
Formula used:
1) Kinetic energy =
2) Work = kinetic energy
Solution:
1)
Kinetic energy is given by,
Kinetic energy =
K.E. =
K.E. = 21.16 Joule
Thus, option (d) is the correct answer.
2)
Work needed to stop the ball is same as the kinetic energy because energy is the capacity to do work. Since, work is opposing the movement of the ball. Thus, Work = - Kinetic energy
Work = -21.16 Joule
Solution :
The angular acceleration, is obtained from the equation of the of rotational motion,
Thus,
or
where is torque, F is force, d is moment arm distance, I is the moment of inertia
Thus,
Now if the force and the moment arm distance are constant, then the
That is when, F = d = constant, then .
Thus, moment of inertia, I is proportional to mass of the bar.
The mass is less for the bar in case (1) in comparison with that with the bar in case (2) due to the holes that is made in the bar.
Therefore, the bar in case (1), has less moment of inertia and a greater angular acceleration.
Answer:
a) λ = 435 nm
, c) c) λ = 4052 nm, d) λ= 95 nm
Explanation:
A) To carry out this excitation, the energy of the laser must be greater than or equal to the energy of the transition of the hydrogen atom, whose states of energy are described by the Bohr model.
En = -13,606 / n² [eV]
therefore the energy of the transition is
ΔE = E₅ -E₂
ΔE = 13.606 (1 / n₂² - 1 / n₅²)
ΔE = 13.606 (1/2² - 1/5²)
ΔE = 2,85726 eV
now let's use Planck's equation
E = h f
the speed of light is related to wavelength and frequencies
c = λ f
f = c /λ
E = h c /λ
λ = h c / E
let's reduce the energy to the SI system
E = 2,85726 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.5716 10⁻¹⁹ J
let's calculate
λ = 6,626 10⁻³⁴ 3 10⁸ / 4,5716 10⁻¹⁹
λ = 4.348 10⁺⁷ m (10⁹ nm / 1 m)
λ = 435 nm
B) photon emission processes from this state with n = 5 to the base state n = 1, can give transition
initial state n = 5
final state n = 4
ΔE = 13.606 (1/4² - 1/5²)
ΔE = 0.306 eV
λ = h c / E
λ = 4052 nm
n = 5
final ΔE (eV) λ (nm)
level
4 0.306 4052
3 0.9675 1281
2 2,857 435
1 13.06 95
n = 4
3 0.661 1876
2 2,551 486
1 11,905 104
n = 3
2 1.89 656
1 12.09 102.5
n = 2
1 10.20 121.6
c) λ = 4052 nm
d) λ= 95 nm