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tankabanditka [31]

2 years ago

13

Which device would you use to most easily measure the illuminance on a surface?

2 answers:

lubasha [3.4K]2 years ago

7 0

A light meter, because illu is a root for light

vladimir2022 [97]2 years ago

3 0

Which device would you use to most easily measure the illuminance on a surface?

A. Diaphragm

B. Photometer

C. Polarizer

D. Light meter

THE CORRECT ANSWER IS D.

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A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <

Svet_ta [14]

Answer:

Hence the answer is E inside $= KQr_{1} /R^{3}$ .

Explanation:

E inside $= KQr_{1} /R^{3}$

so if r1 will be the same then

E $\begin{bmatrix}Blank Equation\end{bmatrix}$ proportional to 1/R3

so if R become 2R

E becomes 1/8 of the initial electric field.

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2 years ago

Read 2 more answers

Lucifer is sitting in a tree that is 15 m above the ground. She jumps straight off the tree in the horizontal direction and land

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2 years ago

Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len

Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

4 0

2 years ago

If a box is pulled with a force of 100 N at an angle of 25

love history [14]

The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

<u>Given the following data:</u>

Force = 100 Newton.

Angle of inclination = 25°

To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

$F_x = Fcos \theta\\\\F_x =100 \times cos25\\\\F_x =100 \times 0.9063$

Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

$F_y = Fsin \theta\\\\F_y =100 \times sin25\\\\F_y =100 \times 0.4226$

Fy = 42.26 Newton.

Read more on horizontal component here: brainly.com/question/4080400

6 0

2 years ago

Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°

lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía ( $S$ ), medida en joules por Kelvin, tenemos la siguiente expresión:

$dS = \frac{\delta Q}{T}$ (1)

Donde:

$Q$ - Ganancia de calor, en joules.

$T$ - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

$dS = \frac{L_{v}}{T}\cdot dm$ (1b)

Donde:

$m$ - Masa del sistema, en kilogramos.

$L_{v}$ - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

$\Delta S = \frac{\Delta m \cdot L_{v}}{T}$

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que $m = 0.50\,kg$ , $L_{v} = 2.7\times 10^{5}\,\frac{J}{kg}$ and $T = 630.15\,K$ , entonces el cambio de entropía es:

$\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}$

$\Delta S = 214.235 \,\frac{J}{K}$

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0

2 years ago