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aev [14]
4 years ago
7

No wrong answers please

Physics
1 answer:
zepelin [54]4 years ago
4 0
Sorry i can’t help sorry good luck tho!!!!
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A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.
SOVA2 [1]

Answer:

The density of the mixture is 0.55kg/m^3

Explanation:

P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole

PV = nRT

V = nRT/P = 1×8.314×273/100 = 22.70m^3

Mass of N2 = 0.4×28 = 11.2kg

Mass of H2 = 0.6×2 = 1.2kg

Mass of mixture = 11.2 + 1.2 = 12.4kg

Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3

3 0
4 years ago
A rubber ball with a mass of 0.145 kg is dropped from rest. From what height (in m) was the ball dropped, if the magnitude of th
Elena-2011 [213]

Answer:

1.55 m

Explanation:

Momentum: This can be defined as the product of  mass of a body and it velocity. the S.I unit of momentum is kgm/s.

Mathematically,

Momentum can be represented as,

M = mv................................. Equation 1

Where m = mass of the body, v = velocity of the body, M = momentum.

Making v the subject of the equation,

v = M/m........................................... Equation 2

Given: M = 0.80 kg.m/s, m = 0.145 kg.

Substituting into equation 2,

v = 0.8/0.145

v = 5.52 m/s.

Using the equation of motion,

v² = u² + 2gs ....................... Equation 3.

Where v = final velocity of the rubber ball, u = initial velocity of the rubber ball, s = distance, g = acceleration due to gravity.

Given: v = 5.52 m/s, u = 0 m/s, g = 9.81 m/s².

Substituting into equation 2

5.52² = 0² + 2(9.81)s

30.47 = 19.62s

s = 30.47/19.62

s = 1.55 m.

Thus the ball was dropped from a height of 1.55 m

8 0
4 years ago
Which of the following is an example of the law of conversion of energy
ruslelena [56]
Is there a picture so I can help you
8 0
3 years ago
If the Greek got Zeus is the ruler of the universe will we ever see him
pav-90 [236]

Answer:

no

Explanation:

4 0
3 years ago
Read 2 more answers
El espectro visible en el aire está comprendido entre la longitud de onda 450 nm del color azul, Determina la velocidad de propa
TiliK225 [7]

Answer:

v = 2,99913 10⁸ m / s

Explanation:

The velocity of propagation of a wave is

         v = λ f

in the case of an electromagnetic wave in a vacuum the speed that speed of light

         v = c

When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium

         λ = λ₀ / n

where n is the index of refraction of the material medium.

Therefore the expression is

           v = \frac{\lambda_o}{n} f

Let's look for the frequency of blue light in a vacuum

           f =\frac{c }{\lambda_o}  

           f = \frac{3 \ 10^8}{450 \ 10^{-9}}

           f = 6.667 10¹⁴ Hz

the refractive index of air is tabulated

           n = 1,00029

let's calculate

           v = \frac{450 \ 10^{-9} }{1.00029}  \ 6.667 \ 10^{14}450 10-9 / 1,00029 6,667 1014

            v = 2,99913 10⁸ m / s

we can see that the decrease in speed is very small

8 0
3 years ago
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