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3 years ago

1.- Se desea elevar un cuerpo de 1500kg utilizando una elevadora hidráulica de plato grande

Physics

1 answer:

kvv77 [185]3 years ago

5 0

Answer:

181.48 N

Explanation:

Calculate the area :

Area = pi * r² ;

pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m

Area 1, A1 = 3.14 * 0.1² = 0.0314 m²

Area 2, A2 = 3.14 * 0.9² = 2.5434 m²

Force, F = mass * acceleration due to gravity

F2 = 1500 * 9.8 = 14700 N

Force 1 / Area 1 = Force 2 / Area 2

Force 1 = (Force 2 / Area 2), * Area 1

Force 1 = (14700 / 2.5434) * 0.0314

Force = 5779.6650 * 0.0314

= 181.48 N

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Small cubes that are 10 cm on a side and larger ones that are 12 cm on a side are submerged in water. Cubes A and B are made of

Digiron [165]

Answer

given,

small cube side = 10 cm

larger cube side = 12 cm

density of steel = 7 g/cm³

density of aluminium = 2.7 g/cm³

density of the water (ρ₁)= 1 g/cm³

Cube A and B made of steel

buoyant force of Cube A

B₁ = ρ₁ V g = 1 x 10 x 10 x 10 x g= 1000 g

for cube B

B₂ = ρ₁ V g = 1 x 12 x 12 x 12 x g= 1728 g

buoyant force of Cube C

B₃ = ρ₁ V g = 1 x 10 x 10 x 10 x g= 1000 g

for cube D

B₄ = ρ₁ V g = 1 x 12 x 12 x 12 x g= 1728 g

buoyant force acting on the cube depends on the density of the fluid

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B₂ = B₄ > B₁ = B₃

8 0

3 years ago

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The pl

Aleks04 [339]

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

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5 0

3 years ago

Does wavelength affect the energy of a wave?

Galina-37 [17]

Answer:

Not quite

Explanation:

The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength

What determines the strength of a wave?

Wave height is affected by wind speed, wind duration (or how long the wind blows), and fetch, which is the distance over water that the wind blows in a single direction. If wind speed is slow, only small waves result, regardless of wind duration or fetch.

So,

As Wavelength increases, The energy of the wave spreads and it decreases

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2 years ago

A parallel-plate capacitor consists of plates of area 1.5 x 10^-4 m^2 separated by 2.0 mm The capacitor is connected to a 12-V b

Katen [24]

Answer:

4.78 x 10^-11 J

Explanation:

A = 1.5 x 10^-4 m^2

d = 2 mm = 2 x 10^-3 m

V = 12 V

Let C be the capacitance of the capacitor

C = ε0 A / d

C = (8.854 x 10^-12 x 1.5 x 10^-4) / (2 x 10^-3)

C = 6.64 x 10^-13 F

Energy stored, U = 1/2 CV^2

U = 0.5 x 6.64 x 10^-13 x 12 x 12

U = 4.78 x 10^-11 J

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3 years ago

The metal gold crystallizes in a face centered cubic unit cell with one atom per lattice point. When X-rays with λ = 1.436 Å are

Umnica [9.8K]

Answer:

r = 1.45 Å

Explanation:

given,

λ = 1.436 Å

θ = 20.62°

d = a

n = 2

metal gold crystallizes in a face centered cubic unit cell

Radius of the gold atom = ?

using Bragg's Law

n λ = 2 d sin θ

2 x 1.436 Å = 2 a sin 20.62°

a = 4.077 Å

We know relation of radius for face centered cubic unit cell

$a = \dfrac{4r}{\sqrt{2}}$

$4.077= \dfrac{4\times r}{\sqrt{2}}$

r = 1.45 Å

the radius of a(n) gold atom. is equal to 1.45 Å

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3 years ago