High energy waves have Gamma rays
Answer:
(a) 1462.38 m/s
(b) 2068.13 m/s
Explanation:
(a)
The Kinetic energy of the atom can be given as:
K.E = (3/2)KT
where,
K = Boltzman's Constant = 1.38 x 10⁻²³ J/k
K.E = Kinetic Energy of atoms = 343 K
T = absolute temperature of atoms
The K.E is also given as:
K.E = (1/2)mv²
Comparing both equations:
(1/2)mv² = (3/2)KT
v² = 3KT/m
v = √[3KT/m]
where,
m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg
v = RMS Speed of Helium Atoms = ?
Therefore,
v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 1462.38 m/s</u>
(b)
For double temperature:
T = 2 x 343 K = 686 K
all other data remains same:
v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 2068.13 m/s</u>
Answer:
in oil film λ = 303.57 10⁻⁹ m
in the water film λ = 319.55 10⁻⁹ m
Explanation:
When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,
when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship
v = λ f
in the void we have
c = λ₀ f
we divide the two expression
c / v = λ₀ / λ
the refractive index is
n = c / v
n = λ₀ /λ
λ = λ₀ / n
let's calculate
in oil film
λ = 425 10⁻⁹ / 1.40
λ = 303.57 10⁻⁹ m
in the water film
λ = 425 10⁻⁹ / 1.33
λ = 319.55 10⁻⁹
those wavelengths are in the ultraviolet
Answer:
v = 4.76 m/s
Explanation:
Given,
The distance traveled by her bike, d = 10 miles
The time of her travel, t = 2.1 m/s
The velocity of an object is defined as the distance traveled by the object to the time of travel. Therefore,
V = d/t m/s
= 10 / 2.1
= 4.76 m/s
Hence, The velocity of her bike is, V = 4.76 m/s
Illluminance is the measurement of photo-metric power. That means, illuminance is the rate of photo-metric flux that is received by a surface per area. It is usually expressed as a unit of W/m^2. Thus, from the choices, the answer we're looking for is illuminance.
