Answer:
El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Explanation:
El gasto es el flujo volumétrico de gasolina (
), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:
(1)
Donde:
- Diámetro de la manguera, medido en pies.
- Velocidad medida de salida, medida en pies por segundo.
Si sabemos que
y
, entonces el gasto de gasolina es:


El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Answer:
1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K
Explanation:
The relationship between temperature and thermal energy is
E = K T
The relationship of the speed of light
c =λ f = f / ν 1/λ= ν
The Planck equation is
E = h f
Let's start the transformations
c = f λ = f / ν
f = c ν
E = h f
E = h c ν
E = KT
h c ν = K T
T = h c ν / K =( h c / K) ν
Let's replace the constants
h = 6.63 10⁻³⁴ J s
c = 3 10⁸ m / s
K = 1.38 10⁻²³ J / K
v = 1 cm-1 (100 cm / 1 m) = 10² m-1
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²
A = h c / K = 1,441 10⁻²
T = 1.44K
ν = 103 cm⁻¹ = 103 10² m
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²
T = 148K
1 Rydberg = 1.097 10 7 m
As we saw at the beginning the λ=1 / v
T = (h c / K) 1 /λ
T = 1,441 10⁻² 1 / 1,097 10⁷
T = 1.3 10⁻⁹ K
E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J
E = KT
T = E/K
T = 1.6 10⁻¹⁹ /1.38 10⁻²³
T = 1.16 10⁴ K
Answer:
The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Explanation:
As data is incomplete here, so by seeing the complete question from the search the data is
vx_0=1.1 x 10^6
ax=0 As acceleration is zero in the horizontal axis so
Equation of motion in horizontal direction is given as


Now for the vertical distance
vy_o=0
than the equation of motion becomes

Now using this acceleration the value of electric field is calculated as

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
The energy stored in a capacitor is
E = (1/2) · (capacitance) · (voltage)²
E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²
E = (3 x 10⁻⁶ F) · (144 V²)
<em>E = 4.32 x 10⁻⁴ Joule</em>
(That's 0.000432 of a Joule)
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