You read that light waves travel through air and water at different speeds. What is one way this can affect what you see around

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

Therefore, its weight in air is noted as

Next, the buoyant force acting on the lead is defined as

We know that

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

Response:

(a)

(b)

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

The moles are calculated based on the provided mass of argon:

Consequently, the work amounts to:

Best regards.

Answer:

0.000047N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

Thus, the power absorbed can be calculated as:

Power = Intensity x Area

This equals = 1.4 x 10^3 x(10)

Thus,

14000 Watts = 14 kWatt

However, the radiation pressure can be defined as

time-averaged intensity divided by the speed of light in a vacuum

So,

P = (1.4 x 1000)/c

Also,

F= P x A

Thus,

((1.4 x 1000)/(3 x10^8)) x 10

This results in

=0.000046666N

Rounded to two significant figures gives us

=0.000047 N

Respuesta:

Explicación:

Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!