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2 years ago

What force holds the lattice structure in place in a crystal?

2 answers:

4 0

At the end it depends on the type of crystal.

There are ionic crystals (like sodium chloride), covalent cristals (like diamond) and metallic crystals (like aluminum).

The context of the question may tell you if it is referred to one type of crystal or other.

I gues it may be about salt compounds, which are ionic crystals. In this case, the force that holds the lattice structure in place in a crystal is ionic (electrical attraction between cations and anions).

In the other two cases, the force is covalent bond in one case and metallic bond in the other case.

MAXImum [283]2 years ago

3 0

The attraction between ions of opposite charge.

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A certain microwave has a

Tatiana [17]

Answer:

$9.375\times 10^{9} Hz$

Explanation:

Frequency is mathematically defined as the quotient of speed divided by wavelength.

$Frequency= \frac {v}{W}$

where

v-is the speed of light

-w is wavelength.

Given the speed of the wave as $V=300000000 \ m/s$ and the wavelength $\lambda= 0.032\ m$ , we substitute these values in the Frequency function to solve for frequency:

$Frequency=\frac {300000000}{0.032}=9375000000 Hz=9.375\times 10^{9} Hz$

Hence, the wave's frequency is $9.375\times 10^9\ Hz$

5 0

3 years ago

12. What is the net force on a 1-kg apple held at rest in your hand?

Fudgin [204]

The apple is not accelerating. That's a big tip off that the net force acting on it is zero.

3 0

2 years ago

A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

4 0

3 years ago

Anyone want to help a brother out

s344n2d4d5 [400]

I would say B but I’m not 100%

5 0

2 years ago

Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object

ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.

Can you solve this system of equations seeing them like this, or do you need more help?

6 0

2 years ago