A thin rod rotates at a constant angular speed. In case A the axis of rotation is perpendicular to the rod at its center. In cas
e B the axis is perpendicular to the rod at one end. In which case, if either, are there points on the rod that have the same tangential speeds?
1 answer:
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Answer:
So the two wavelength range will over lap
Explanation:
The Rydberg equation is given by
m is the mass of electron
k = 1/4π∈₀
∈₀ = is the permitivity of free space
e is the charge of electron
h is the plank constant
c is the speed of light in vaccum
z is the atomic number = 1
where R is the Rydberg constant = 1.097373 × 10⁷m⁻¹
For Paschen series of H spectrum
in Paschen series of H spectrum
The maximum wavelength occur for
The minimum wavelength occur for = ∞
The brackett series of H spectrum
The maximum wavelength occur for
The minimum wavelength occur for = ∞
So the two wavelength range will over lap
Answer:
a) v = 11.24 m / s , θ = 17.76º b) Kf / K₀ = 0.4380
Explanation:
a) This is an exercise in collisions, therefore the conservation of the moment must be used
Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved
Recall that moment is a vector quantity so it must be kept on each axis
X axis
initial moment. Before the crash
p₀ₓ = m₁ v₁
where v₁ = -25.00 me / s
the negative sign is because it is moving west and m₁ = 900 kg
final moment. After the crash
= (m₁ + m₂) vx
p₀ₓ = p_{x f}
m₁ v₁ = (m₁ + m₂) vₓ
vₓ = m1 / (m₁ + m₂) v₁
let's calculate
vₓ = - 900 / (900 + 1200) 25
vₓ = - 10.7 m / s
Axis y
initial moment
= m₂ v₂
where v₂ = - 6.00 m / s
the sign indicates that it is moving to the South
final moment
p_{fy}= (m₁ + m₂)
p_{oy} = p_{fy}
m₂ v₂ = (m₁ + m₂) v_{y}
v_{y} = m₂ / (m₁ + m₂) v₂
we calculate
= 1200 / (900+ 1200) 6
= - 3,428 m / s
for the velocity module we use the Pythagorean theorem
v = √ (vₓ² + v_{y}²)
v = RA (10.7²2 + 3,428²2)
v = 11.24 m / s
now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)
tan θ = / Vₓ
θ = tan-1 v_{y} / vₓ)
θ = tan -1 (3,428 / 10.7)
θ = 17.76º
This angle is from the west to the south, that is, in the third quadrant.
b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship
Kf = 1/2 (m1 + m2) v2
K₀ = ½ m₁ v₁² + ½ m₂ v₂²
Kf = ½ (900 + 1200) 11.24 2
Kf = 1.3265 105 J
K₀ = ½ 900 25² + ½ 1200 6²
K₀ = 2,8125 10⁵ + 2,16 10₅4
K₀ = 3.0285 105J
the wasted energy is
Kf / K₀ = 1.3265 105 / 3.0285 105
Kf / K₀ = 0.4380
this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy