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2 years ago

A person is standing outdoors in the shade where the temperature is 35 °C.

Physics

1 answer:

Free_Kalibri [48]2 years ago

6 0

Answer:

(a) Eₐ = 6.36 J/s

(b) Eₐ = 4.64 J/s

Explanation:

Stefan-Boltzmann law: States that the total energy per second radiated or absorbed by a black body is directly proportional to the absolute temperature.

Using, Stefan-Boltzmann equation

Eₐ =eσAT⁴ ................ Equation 1

where Eₐ = Radiant energy absorbed per seconds, e = emissivity, σ = stefan - boltzman constant, A = Surface area. and T = temperature in kelvin

(a) Where e = 0.89, σ = 5.67 ×10⁻⁸ watt/m²/K⁴, A = 140 cm² = 140 cm²(m²/10000cm²) = 0.014 m², T = 35 °C = (35 + 273) K = 308 K.

Applying these values in equation 1 above,

Eₐ = 0.89 × 5.67 ×10⁻⁸ × 0.014 × (308)⁴

Eₐ =6.36 J/s

(b) when e = 0.65,

∴ Eₐ = 0.65 × 5.67 × 10⁻⁸ × 0.014 × (308)⁴

Eₐ = 4.64 J/s

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If the humidity in a room of volume 450 m3 at 30 ∘C is 75%, what mass of water can still evaporate from an open pan?

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From tables,

SVP at 30°C = 4.24 kPa

From ideal gas expressions;
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2 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

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2 years ago

The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan

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Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴ × 1000 m

r = 1.08 × 10⁷ m, G = 6.67 × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08 × 10⁷ )²/75( 6.67 × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

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2 years ago

A mover uses a ramp to load a crate of nails onto a truck. The crate, which must be lifted 1.5 m from the street to the bed of t

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Answer: M.A = 3

Explanation:

A ramp is an example of an inclined plain. Where the

Height H = 1.5 m

Length L = 4.5 m

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M.A = L/H

Substitute the values of L and H into the formula

M.A = 4.5/1.5 = 3

The mechanical advantage of the ramp is 3

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2 years ago