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3 years ago

Compare the speed of light in water to the speed of sound in water

1 answer:

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Answer:As a rule sound travels slowest through gases,faster through liquids,and fastest through solids.The speed of light as it travels through air and space is much faster than that of sound; it travels at 300 million meters per second or 273,400 miles per hour.Speed of light in water = 226 million m/s or 205,600 mph.

Explanation:dont have one

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How original were Newton’s contributions to science? (In what ways did Newton depend on the mechanical view?)

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3 years ago

6th grade science please help

katovenus [111]

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3 years ago

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Científico que formuló 3 leyes de astronomía

s2008m [1.1K]

Answer:

El astrónomo alemán Johannes Kepler

Explanation:

Primera Ley:

Los planetas giran alrededor del Sol siguiendo una trayectoria elíptica. El Sol se sitúa en uno de los focos de la elipse.

La excentricidad e de una elipse es una medida de lo alejado que se encuentran los focos del centro.

Pues bien, la mayoría de las órbitas planetarias tienen un valor muy pequeño de excentricidad, es decir e ≈ 0. Esto significa que, a nivel práctico, pueden considerarse círculos descentrados.

Segunda Ley:

La recta que une el planeta con el Sol barre áreas iguales en tiempos iguales.Para que esto se cumpla, la velocidad del planeta debe aumentar a medida que se acerque al Sol. Esto sugiere la presencia de una fuerza que permite al Sol atraer los planetas, tal y como descubrió Newton años más tarde.

Tercera ley de Kepler:

La tercera ley, también conocida como armónica o de los periodos, relaciona los periodos de los planetas, es decir, lo que tardan en completar una vuelta alrededor del Sol, con sus radios medios.

Para un planeta dado, el cuadrado de su periodo orbital es proporcional al cubo de su distancia media al Sol.

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3 years ago

A small object with mass m, charge q, and initial speed v0 = 5.00 * 103 m>s is projected into a uniform electric field betwee

kotykmax [81]

Answer:

q/m = 2177.4 C/kg

Explanation:

We are given;

Initial speed v_o = 5 × 10³ m/s = 5000 m/s

Now, time of travel in electric field is given by;

t_1 = D_1/v_o

Also, deflection down is given by;

d_1 = ½at²

Now,we know that in electric field;

F = ma = qE

Thus, a = qE/m

So;

d_1 = ½ × (qE/m) × (D_1/v_o)²

Velocity gained is;

V_y = (a × t_1) = (qE/m) × (D_1/v_o)

Now, time of flight out of field is given by;

t_2 = D_2/v_o

The deflection due to this is;

d_2 = V_y × t_2

Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)

d_2 = (qE/m) × (D_1•D_2/(v_o)²)

Total deflection down is;

d = d_1 + d_2

d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]

d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]

Making q/m the subject, we have;

q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]

We have;

E = 800 N/C

d = 1.25 cm = 0.0125 m

D_1 = 26.0 cm = 0.26 m

D_2 = 56 cm = 0.56 m

Thus;

q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]

q/m = 312500/143.52

q/m = 2177.4 C/kg

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3 years ago

A sphere of radius 5.00 cmcm carries charge 3.00 nCnC. Calculate the electric-field magnitude at a distance 4.00 cmcm from the c

kogti [31]

Answer:

a) E = 8628.23 N/C

b) E = 7489.785 N/C

Explanation:

a) Given

R = 5.00 cm = 0.05 m

Q = 3.00 nC = 3*10⁻⁹ C

ε₀ = 8.854*10⁻¹² C²/(N*m²)

r = 4.00 cm = 0.04 m

We can apply the equation

E = Qenc/(ε₀*A) (i)

where

Qenc = (Vr/V)*Q

If Vr = (4/3)*π*r³ and V = (4/3)*π*R³

Vr/V = ((4/3)*π*r³)/((4/3)*π*R³) = r³/R³

then

Qenc = (r³/R³)*Q = ((0.04 m)³/(0.05 m)³)*3*10⁻⁹ C = 1.536*10⁻⁹ C

We get A as follows

A = 4*π*r² = 4*π*(0.04 m)² = 0.02 m²

Using the equation (i)

E = (1.536*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.02 m²)

E = 8628.23 N/C

b) We apply the equation

E = Q/(ε₀*A) (ii)

where

r = 0.06 m

A = 4*π*r² = 4*π*(0.06 m)² = 0.045 m²

Using the equation (ii)

E = (3*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.045 m²)

E = 7489.785 N/C

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3 years ago