Answer:
a) The distance of spectator A to the player is 79.2 m
b) The distance of spectator B to the player is 43.9 m
c) The distance between the two spectators is 90.6 m
Explanation:
a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:
x = v * t
where:
x = position of the spectators
v = speed of sound
t = time
Then, the position for spectator A relative to the player is:
x = 343 m/s * 0.231 s = 79.2 m
b)For spectator B:
x = 343 m/s * 0.128 s
x = 43.9 m
The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.
c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:
(Distance AB)² = A² + B²
(Distance AB)² = (79.2 m)² + (43.9 m)²
Distance AB = 90. 6 m
Answer:
a) 0.99999 e
b) 1*10^-5 s
c) 3*10^-2 m
Explanation:
Given data:
Electron energy : 50 GeV
assuming the rest mass of an electron = 0.5 MeV/c^2
A) Determine the velocity of each electron
velocity of each electron ( v ) ≈ 0.999 e
B) How much time does it take the electrons to travel the length of the accelerator given that the accelerator = 3km long
Lo = 3 * 10^3 m
T = (3 * 10^3 )/ (0.999 * 3 * 10^8)
= 1 * 10 ^-5
C) How long does the accelerator look form the perspective of the electrons
L = Lo / r = Lo 
hence L = 3 * 10^-2 m
Answer:
n₁ = 3
Explanation:
The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,
ΔE = 
- E₀ = - k²e² / 2m (1 / 
²2 - 1 / n₀²)
The energy of this transition is given by the Planck equation
E = h f = h c / λ
h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)
1 / λ = Ry (1/ ² - 1 / n₀²)
Let's apply these equations to our case
λ = 821 nm = 821 10⁻⁹ m
E = h c / λ
E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹
E = 2.423 10⁻¹⁹ J
Now we can use the Bohr equation
Let's reduce to eV
E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV
- E₀ = -13.606 (1 / ² - 1 / n₀²) [eV]
Let's look for the energy of some levels
n 
(eV) - E
(eV)
1 -13,606 E₂-E₁ = 10.20
2 -3.4015 E₃-E₂ = 1.89
3 -1.512 E₄- E₃ = 0.662
4 -0.850375
We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value
Answer:
Explanation:
From the information given:
The diameter of the cylindrical heater (d) = 1 cm
The length of the cylindrical heater (l) = 0.25 m
The ambient air temperature 
= 25° C= (273+25)K = 298 K
The convective heat transfer coefficient (h) = 25 W/m² °C
The electric input Q = 5W
As stated in the question that if radiation is being neglected:-
Let also assume that;
the heat transfer takes place at a steady-state
1-D flow takes place
No external heat generation; &
No force convection takes place;
Then; the heat transfer through the convection can be calculated as:
By solving the above calculation:
T ( surface temperature of the heater) = 50.46° C 122.83° F
Answer:
Explanation:
Wall reactions are never constant as the ladder angle decreases from vertical.
I will ASSUME that the MAXIMUM wall reaction is 40 N before slippage occurs.
Let θ be the ladder angle to horizontal
Moments about the ladder foot will sum to zero
40[5sinθ) - 100[(5/2)cosθ = 0
40[5sinθ) = 100[(5/2)cosθ
200sinθ = 250cosθ
sinθ/cosθ = 250/200
tanθ = 1.25
θ = 51.34019174...
θ = 51°
