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3 years ago

Intermolecular forces hold together which of the following?

Physics

2 answers:

gregori [183]3 years ago

6 0

Answer: The correct answer is molecules.

Explanation:

The intermolecular forces help to hold the molecules together. The force of attraction between the molecules is called intermolecular forces.

In the solid state, the molecules are tightly packed. There is a strong force of attraction between the molecules.

In the liquid state, the molecules are loosely packed in comparison to the solid state.

In the gaseous state, the molecules are more loosely in comparison to the liquid state. There is a weak force of attraction between the molecules.

Therefore, intermolecular forces hold together molecules.

andrezito [222]3 years ago

4 0

Intermolecular forces exist between the molecules of a substance.

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When is a white dwarf formed?

Anna [14]

When a red giant has insufficient mass it pretty much sheds its outer layer to leave the center core know as the white dwarf

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3 years ago

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WHAT DOES DENSITY HAVE TO DO WITH PLATE TECTONICS?<br> Explain

Korvikt [17]

Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.

Explanation:

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3 years ago

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las

topjm [15]

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

$w=\frac{2.44 \lambda L}{D}$

where $\lambda$ is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

$L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m$

4 0

3 years ago

A model engine accelerated forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates coward at

Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

a₂ = 16 m/s²

t₂ = 5.0 s

a₃ = -12 m/s²

v₃ = 0 m/s

Find: t

First, find v₁:

v₁ = a₁t₁ + v₀

v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)

v₁ = 30 m/s

Next, find v₂:

v₂ = a₂t₂ + v₁

v₂ = (16 m/s²) (5.0 s) + (30 m/s)

v₂ = 110 m/s

Finally, find t₃:

v₃ = a₃t₃ + v₂

(0 m/s) = (-12 m/s²) t₃ + (110 m/s)

t₃ = 9.2 s

The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0

3 years ago

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final:

$A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]$

$A_{f} = 5102.752\,cm^{2}$

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0

3 years ago