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:
h = height of the hotel room from the ground floor = 237.4m
Explanation:
Change in Potential Energy of tourist = ΔPE = PE2 – PE1 = mgh
PE1 is the potential energy of tourist at the ground floor
PE1 is the potential energy of tourist at the top (hotel room)
Given
PE1 = − 2.01 × 10⁵ J
PE2 = 0J
PE2 – PE1 = mgh
0 – (− 2.01 × 10⁵ J) = mgh
2.01 × 10⁵ J = 86.4×9.8×h
h = 2.01 × 10⁵/(86.4×9.8) = 237.4m
Answer:
The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
Explanation:
P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.
Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.
Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
Answer:
The first graph
Explanation:
Graph A shows acceleration.
V: velocity of wave
f: frequency
L: wavelenght
v = fL => L = v/f => L = (3x10^8)/(900x10^3) => L = 3.33 x 10^2m
