Speed = (wavelength) x (frequency
Speed = (.020 m) x (5 / sec)
Speed = 0.1 m/s
Answer:
What is the average translational kinetic energy of molecules in an ideal gas at 37°C? The average translational energy of a molecule is given by the equipartition theorem as, E = 3kT 2 where k is the Boltzmann constant and T is the absolute temperature.
Explanation:
The average translational energy of a molecule is given by the equipartition theorem as, E = 3kT 2 where k is the Boltzmann constant and T is the absolute temperature.
The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i = 
where R is the resistance
R = 
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i = 
q = 
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
1) Data:
Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m
2) Questions: V at Y = 3m and X at Y = 3 m
3) Calculate components of the initial velocity
Vox = Vo * cos(37°) = 15.97 m/s
Voy = Vo * sin(37°) = 12.04 m/s
4) Formulas
Vx = constant = 15.97 m/s
X = Vx * t
Vy = Voy - g*t
Y = Yo + Voy * t - g (t^2) / 2
5) Calculate t when Y = 3m (first time)
Use g ≈ 9.8 m/s^2
3 = 12.04 * t - 4.9 t^2
=> 4.9 t^2 - 12.04t + 3 = 0
Use the quadratic equation to solve the equation
=> t = 0.28 s and t = 2.18s
First time => t = 0.28 s.
6) Calculate Vy when t = 0.28 s
Vy = 12.04 m/s - 9.8 * 0.28s = 9.3 m/s
7) Calculate V:
V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s
tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°
Answer: V ≈ 18.5 m/s, with angle ≈ 30°
8) Calculate X at t = 0.28s
X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m
Answer: X ≈ 4,5 m
Answer:
Option D
Explanation:
The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.
In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.
In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.
