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Sphinxa [80]
3 years ago
13

A box has a mass of 150kg and the surface area of the bottom of the box is :

Physics
1 answer:
charle [14.2K]3 years ago
3 0

Answer:

Pressure = 100 Pascals

Explanation:

pressure =  \frac{force}{area}  \\  =  \frac{150 \times 10}{15}  \\  =  \frac{1500}{15}  \\  = 100 pa

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Problema 1. Un Clavadista se lanza desde un trampolín a diferentes alturas 10 m, 3 m, y 1 m, Calcular:
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La velocidad del buzo es de 7.04 m/s, 8.84 m/s y 2.27 m/s respectivamente.

<h3 /><h3>Velocidad </h3>

La velocidad es la relación entre la distancia total recorrida y el tiempo total empleado. Está dado por:

Velocidad = distancia/tiempo

Para 10m de altura:

  • Velocidad = 10 m/1.42 s = 7.04 m/s

Para 3m de altura:

  • Velocidad = 3 m/0.78 s = 3.84 m/s

Para 1m de altura:

  • Velocidad = 1 m/0.44 s = 2.27 m/s

La velocidad del buzo es de 7.04 m/s, 8.84 m/s y 2.27 m/s respectivamente.

Obtenga más información sobre la velocidad en: brainly.com/question/4931057

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2 years ago
HEEEEELLLPPPPPP!!!!!!!!!!! PPPPPPPPPPPLLLEAASSEE!!!
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Answer:

The arrows always start at the magnet's north pole and point towards its south pole. When two like-poles point together, the arrows from the two magnets point in OPPOSITE directions and the field lines cannot join up. So the magnets will push apart (repel).

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3 years ago
Scientists have discovered just over _____ different elements with unique properties.
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100. (: hope this helps
8 0
2 years ago
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What kind of energy does a rubber band have when it is stretched?
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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
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