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BlackZzzverrR [31]
2 years ago
9

You are standing in front of a pool of water and see your face in the water but when a stone is dropped into the pool you no mor

e see your face
Explain why please ​
Physics
1 answer:
Wewaii [24]2 years ago
4 0

Answer:

• Before, the water is calm with a regular reflection surface. So, all the light rays from your face will be reflected directly to your eyes hence viewing your face clearly.

• After throwing a stone in the pool, the water distorts or disorganises due to change in volume and molecular dislocation. So, it forms an irregular reflecting surface. Hence, all light rays from the face are reflected in different directions hence you can't see your face clearly.

Explanation:

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Compare the gravitational acceleration on the following objects compared to the Sun using:
arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765  x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0
2 years ago
The stair stepper is a novel exercise machine that attempts to reproduce the work done against gravity by walking up stairs. Wit
nika2105 [10]

Answer:

The total work done by Brad each day is 176400 J

Explanation:

Hi there!  The work done by a force (F) pointed in the same direction as the displacement (d) is calculated as follows:

W = F · d

The force applied is equal to the weight of Brad, that is calculated as follows:

Weight = m · g

Where:

m =  mass of Brad

g = acceleration due to gravity (9.8 m/s²)

Then:

Weight = 60 kg · 9.8 m/s² = 588 N

 

Let´s find the vertical distance traveled by Brad each day:

He exercises 20 min per day. Each minute Brad does 60 steps. In total, Brad steps up (20 min · 60 steps/min) 1200 steps. If each step has a height of 0.25 m, the total distance traveled by Brad will be

(1200 steps · 0.25 m/step) 300 m.

Then, the total work done by Brad is

W = F · d

W =  588 N · 300 m

W = 176400 J

The total work done by Brad each day is 176400 J

6 0
3 years ago
Why are red stars cooler than blue stars?
asambeis [7]
OK both stars are burning and as they burn they emit electromagnetic radiation (light) and as we move in the light spectrum red comes before blue, so the star that looks red is cooler because as the temperature decreases the colour decreases
6 0
3 years ago
Read 2 more answers
Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p
boyakko [2]

Answer:

The number of free electrons per cubic meter is 7.61\times 10^{28}\ m^{-3}

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, \sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}

Density of metal, \rho=10.5\ g/cm^3

Atomic weight, A = 107.87 g/mol

Let n is the number of  free electrons per cubic meter such that,

n=1.3\ N

n=1.3(\dfrac{\rho N_A}{A})

Where

\rho is the density of silver atom

N_A is the Avogadro number

A is the atomic weight of silver

n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})

n=7.61\times 10^{22}\ cm^{-3}

or

n=7.61\times 10^{28}\ m^{-3}

Hence, this is the required solution.

6 0
3 years ago
El Sol está en promedio a 93 millones de millas de la Tierra. ¿A cuántos metros equivale esto? Expréselo usando potencias de die
mars1129 [50]
Hola!

93 millones de millas equivalen a 1,50 x 10¹¹ metros. 

Para calcular este valor, necesitamos saber la equivalencia entre millas y metros (1 milla=1609,34m), y aplicar el siguiente factor de conversión. Para expresar el resultado en potencias de 10, se debe correr la coma a la izquierda hasta obtener un valor de una sola unidad y multiplicar este valor por 10 elevado a la cantidad de espacios que la coma se tuvo que correr a la izquierda:

93000000millas*\frac{1609,34 metros}{1milla}=150000000000,0 metros=1,50* 10^{11} metros

Saludos!
5 0
3 years ago
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