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2 years ago

Intermolecular forces hold together which of the following?

Physics

2 answers:

gregori [183]2 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]2 years ago

4 0

Intermolecular forces exist between the molecules of a substance.

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A tuning fork generates sound waves with a frequency of 240 Hz. The waves travel in opposite directions along a hallway, are ref

Bingel [31]

Answer:

The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

Explanation:

Given that,

Frequency of sound wave = 240 Hz

Distance = 46.0 m

Distance of fork = 14 .0 m

We need to calculate the path difference

Using formula of path difference

$\Delta x=2(L_{2}-L_{1})$

Put the value into the formula

$\Delta x =2((46.0-14.0)-14.0)$

$\Delta x=36\ m$

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{v}{f}$

Put the value into the formula

$\lambda=\dfrac{343}{240}$

$\lambda=1.42\ m$

We need to calculate the phase difference

Using formula of the phase difference

$\phi=\dfrac{2\pi}{\lambda}\times \delta x$

Put the value into the formula

$\phi=\dfrac{2\pi}{1.42}\times36$

$\phi=159.29\ rad$

$\phi\approx 68.2^{\circ}$

Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

7 0

3 years ago

Desde el balcón de un edificio se deja caer una manzana y llega a la planta baja en 5 s. ¿Desde qué piso se dejó caer, si cada p

miv72 [106K]

Answer: 42

Explanation:

I will answer this in English.

We know that the apple needs 5 seconds to reach the ground.

Each floor of the building has a height of 2.88m.

Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:

a = -g

for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:

v = -g*t

for the position we integrate again, now we have an initial height H, so the position is:

p = (-g/2)*t^2 + H

now the apple hits the ground when p = 0, so we can solve this equation to find H.

i will use g = 9.8m/s^2

0 = (-4.9m/s^2)*(5s)^2 + H

H = 122.5 m

now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.

N = 122.5m/2.88m = 42.5

this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)

5 0

2 years ago

Maggie's truck decelerates from 18.00 m/s to rest in 3.30 s. If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculat

Free_Kalibri [48]

Answer:

look at my Explanation

Explanation:

If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculate the magnitude of the net (unbalanced) force that can cause the acceleration.

8 0

2 years ago

Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al

Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire $(m) = 1.10g = 1.10*10^{-3} kg$

Density $(\rho)= 8.92*10^3kg/m^3$

Resistively of copper $(\gamma) = 1.7*10^{-8}\Omega \cdot m$

Resistance (R) = 0.390\Omega

Volume is defined as,

$V= lA \text{ and }\frac{m}{\rho}$

$lA= \frac{1.10*10^{-3}}{8.92*10^3}$

$lA = 1.233*10^{-7} m^3$ (1)

We know that,

$\frac{l}{A} = \frac{R}{\gamma}$

$\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}$

$\frac{l}{A} = 2.2941*10^7 m^{-1}$ (2)

Multiplying equation we have

$l^2 = (1.233*10^{-7})( 2.2941*10^7)$

$l^2 = 2.8286m^2$

$l =\sqrt{2.8286m^2}$

$l = 1.68m$

Therefore the length of the wire is 1.68m

6 0

2 years ago

Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga

Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

6 0

3 years ago