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Nostrana [21]
2 years ago
13

What is the melting point of sand

Physics
1 answer:
telo118 [61]2 years ago
8 0
The kind of heat necessary to transform sand into a liquid state (eventually becoming glass) is much hotter than any sunny day. To make sand melt, you need to heat it to roughly 1700°C (3090°F), which is approximately the same temperature a space shuttle reaches as it re-enters earth's atmosphere.
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Which is required for
vaieri [72.5K]

Answer:

You need (B) Two substances in direct contact with one another to have conduction.

I hope this helped at all.

5 0
3 years ago
A ______ is a network that has all connected devices located in the same physical location.
Dafna11 [192]
<span>The answer is Local Area Network (LAN). Examples of LANs, are those in workplaces, libraries, and universities. The nodes are connected and share common resources over a WiFi, Ethernet network, Token rings, and etcetera. The opposite of LAN is a Wide Area Network (WAN) that covers a wide geographical area.</span>




3 0
3 years ago
Variations in the reactive properties of different organic molecules are most closely associated with _____.
azamat

Answer:

the presence or absence of functional groups

Explanation:

The functional group is the group of atoms that characterize a chemical function and that have well-defined characteristic properties.

In organic chemistry, the functional group is a set of submolecular structures, characterized by a specific elementary connectivity and composition that confers specific chemical reactivity to the molecule that contains them. These structures replace the hydrogen atoms lost by saturated hydrocarbon chains. Aliphatic, or open chain, groups are usually represented generically by R (alkyl radicals), while aromatic ones, or derivatives of benzene, are represented by Ar (aryl radicals).

7 0
3 years ago
Read 2 more answers
A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
Natalija [7]

Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

   The mass  of the ball is  m_b  = 490 \ g  = 0.49 \ kg

   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

    \tau = \frac{m}{a}      

=> \tau = \frac{0.490}{  0.028}    

=> \tau = 17.5 \ s    

4 0
3 years ago
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
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