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3 years ago

What are binary compounds

2 answers:

8 0

In chemistry, a binary compound is something consisting of precisely two elements. ... A few examples of chemical elements are hydrogen, oxygen, and iron. This means that a binary compound will be composed of two different chemical elements. In a binary compound, there may be only one of each element.

madreJ [45]3 years ago

4 0

Answer:

a binary phase or binary compound is a chemical compound containing two different elements. Some binary phase compounds are molecular, e.g. carbon tetrachloride. More typically binary phase refers to extended solids.

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Question 1 Water: Start with the Water tab. Note that light areas represent places where the water is high (crests). Dark areas

Andre45 [30]

Answer:

Looks like a cosine function graph. The wave pattern is transversal waves . The faster the amplitude the higher the wave. The force of the drop hitting the water pushes the water down and out causing waves If the water hits from a higher amplitude the waves raise bigger . When you increase the frequency of the water drops the waves move faster but no bigger. When you increase the frequency of the water drops the wave ripples faster .

Explanation:

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3 years ago

Barker is unloading 20kg bottles of water from this delivery truck when one of the bottles tips over and slides down the truck r

Degger [83]

<span>The overall force that is acting on the bottle is gravity. With the incline being 30 degrees the full force of gravity isn't acting on the bottle becuase the ramp isn't allowing the bottle to go straight down. By taking the sin of 30 degrees you find the proportion of gravity that is acting on the bottle to be 4.9 meters per second and the bottle weights 20 kg so the force acting on the bottle is 98 Newtons.</span>

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3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

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3 years ago

Savvas realize hot on the inside

Liula [17]

Answer:

Thats sounds like a club

Explanation:

8 0

3 years ago

Read 2 more answers

A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motio

QveST [7]

Answer:

(a) 0.31

(b) 0.245

Explanation:

(a)

F' = μ'mg.................... Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg............. Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg.................. Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg.............. Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ = 60/(25×9.8)

μ = 60/245

μ = 0.245

3 0

3 years ago

What are binary compounds​

What are binary compounds