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grin007 [14]
3 years ago
7

What is the pressure on a horizontal surface with an area of 8 msquared that is 4 m​ underwater? Use 1000 kg divided by m cubed

for the density of water and 9.8 m divided by s squared for the acceleration due to gravity. What is the correct relationship to use to find the pressure on the horizontal​ surface?
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
7 0

Answer:

140525 pa

Explanation:

h = depth of water = 4 m

A = Area of surface = 8 m²

\rho = Density of water = 1000 kg m⁻³

tex]g[/tex] = acceleration due to gravity = 9.8 ms⁻²

P_{o} = Atmospheric pressure = 101325 Pa

P = Pressure at the given depth "h"

Pressure at the given depth is given as

P = P_{o} + \rho g h\\P = 101325 + (1000) (9.8) (4)\\P = 140525 pa

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r = inner radius of mug. h = height of coffee = 6 cm. 
Density of coffee = Density of water = 1 g/ml 

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Volume of coffee = πr²h
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When would the carrying capacity of an area be most likely to change? Choose the correct answer.
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Explanation:

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Una carga positiva de 4 x 10-5 C, se encuentra a 0.05 m de otra carga positiva de 2 x 10-5 C. Calcular la fuerza que se ejerce e
KATRIN_1 [288]

Answer:

La fuerza que se ejerce entre las dos cargas es 2880 N.

Explanation:

La ley de Coulomb indica que los cuerpos cargados sufren una fuerza atractiva o repulsiva al acercarse. La fuerza es atractiva si las cargas son del signo opuesto y repulsión si son del mismo signo. El valor de la fuerza es proporcional al producto del valor de sus cargas e inversamente proporcional al cuadrado de la distancia que los separa. Esto se expresa matemáticamente como:

F=k*\frac{Q*q}{r^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. Se mide en Newtons (N).
  • Q y q son los valores de las dos cargas puntuales. Se miden en culombios (C).
  • r es el valor de la distancia que los separa. Se mide en metros (m).
  • k es una constante de proporcionalidad llamada constante de la ley de Coulomb.

En este caso:

  • F= ?
  • Q= 4*10⁻⁵ C
  • q= 2*10⁻⁵ C
  • r= 0.05 m
  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{4*10^{-5} C*2*10^{-5}C }{(0.05 m)^{2} }

F= 2880 N

<u><em>La fuerza que se ejerce entre las dos cargas es 2880 N.</em></u>

7 0
3 years ago
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