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Kryger [21]
3 years ago
6

Late on an autumn day, the relative humidity is 44.0% and the temperature is 20.0°C. What will the relative humidity be that ev

ening when the temperature has dropped to 10.0°C, assuming constant water vapor density?
Physics
1 answer:
tester [92]3 years ago
8 0

Answer:

humidity is 80.95 %

Explanation:

given data

humidity = 44% = 0.44

temperature = 20.0°C

temperature = 10.0°C

to find out

relative humidity

solution

we know humidity formula that is

humidity = \frac{vapor density}{saturated vapor density} × 100   ...................1

so here at 20.0°C

vapor density is

vapor density = humidity × saturated vapor density

here

saturated vapor density at 20.0°C is  17.3 gm/m³

so

vapor density = 0.44 × 17.3

vapor density = 7.61

and

for 10.0°C  

saturated vapor density at 10.0°C is  9.4 gm/m³

so from equation 1

humidity = \frac{vapor density}{saturated vapor density} × 100  

humidity = \frac{7.61}{9.4} × 100  

humidity = 80.95 %

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El coeficiente de variación de la resistencia con la temperatura del carbón es -0.0005/°c.Si la resistencia de una resistencia d
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Answer:

 R (120) = 940Ω

Explanation:

The variation in resistance with temperature is linear in metals

           ΔR (T) = R₀ α ΔT

where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC

let's calculate

            ΔR = 1000 (-0,0005) (120-0)

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            R (120) = -60 + 1000

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3 0
4 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a
marishachu [46]

Answer:

-9.46 N

Explanation:

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This value will be the projection of the force vector, on the x-axis.

For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₁ₓ, and r = F₁

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º-55º = 125º

⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N

We can repeat the process for F₂, as follows:

For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₂ₓ, and r = F₂

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º + 53.3º = 233.3º

⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N

The total component of both forces along the x axis, can be found just adding both components, as follows:

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5 0
3 years ago
A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

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