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2 years ago

What is the coordination number of each atom in Silicon?

Physics

2 answers:

Marrrta [24]2 years ago

6 0

Answer:

the coordination number of silicon is two.

sp2606 [1]2 years ago

3 0

In most silicon-oxygen compounds, the silicon atom is 4-fold coordinated. Exceptionally, silicon is found in 6-fold coordination, as in the mineral thaumasite, which also may be formed in the degradation, and sometimes subsequent weakening, of certain concretes.

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A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

3 years ago

A race car accelerates from 0.00m/s to 40.0m/s with a displacement of 50.0m. What was the cars

Vladimir79 [104]

Answer:

Explanation:

Using the equation of motion v² = u²+2as

v is the final velocity = 40m/s

u is the iniyail velocity 0m/s

a is the acceleration

s is the displacement

Substituting in the formula;

40² = 0²+2a(50)

1600 = 100a

Divide both sides by 100

100a/100 = 1600/100

a = 16

Hence the car acceleration is 16m/s²

7 0

3 years ago

If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?

Alex Ar [27]

Answer:

They have same density

Explanation:

The density of an object is defined as

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

Let's call $m_c$ and $V_c$ the mass and the volume of ball C, respectively. Therefore, the density of ball C is:

$d_c = \frac{m_c}{V_c}$

We know that the volume of ball C is 3 times the volume of ball D, so

$V_c = 3 V_d \rightarrow V_d = \frac{V_c}{3}$

And we also know that ball D has 1/3 the mass of ball C:

$m_d = \frac{m_c}{3}$

So, the density of ball D is:

$v_d = \frac{m_d}{V_d}=\frac{m_c/3}{V_d/3}=\frac{m_c}{V_c}=d_c$

Therefore, the two balls have same density.

6 0

3 years ago

In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin

GuDViN [60]

Answer:

The velocity is $v_b = 20.17 \ m/s$

Explanation:

From the question we are told that

The mass of the ball is $m = 0.245 \ kg$

The radius is $r = 59.8 \ cm = 0.598 \ m$

The force is $F = 30.9 \ N$

The speed of the ball is $v = 16.0 \ m/s.$

Generally the kinetic energy at the top of the circle is mathematically represented as

$K_t = \frac{1}{2} * m * v^2$

=> $K_t = \frac{1}{2} * 0.245 * 16.0 ^2$

=> $K_t = 31.36 \ J$

Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as

$W = F * d$

Here d is the length of a semi - circular arc which is mathematically represented as

$d = \pi * r$

$W = 30.9 * 0.598$

$W = 18.48 \ J$

Generally the kinetic energy at the bottom is mathematically represented as

$K_b = \frac{1}{2} * m * v_b^2$

=> $K_b = \frac{1}{2} * 0.245 * v_b^2$

=> $K_b = 0.1225 * v_b^2$

From the law of energy conservation

$K_t + W =K_b$

=> $31.36+ 18.48 = 0.1225 * v_b^2$

=> $v_b = 20.17 \ m/s$

3 0

3 years ago

The voltage across the input terminals of a transformer is 120 V. The primary has 25 loops and the secondary has 50 loops. The v

saw5 [17]

Answer:

240 V

Explanation:

Vp = 120 V

Np = 25

Ns = 50

Vs = ?

Vs / Vp = Ns / Np

Vs / 120 = 50 / 25

Vs / 120 = 2

Vs = 240 V

5 0

3 years ago