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shtirl [24]
3 years ago
5

adeen says that you can increase the resistance of a copper wire by hammering the wire to make it narrower and longer. Arnell sa

ys that you can increase its resistance by heating the wire. Which one, if either, is correct, and why
Physics
1 answer:
ludmilkaskok [199]3 years ago
7 0

Answer:

<em>Both of Arnell and Adeen are right.</em>

<em></em>

Explanation:

For Adeen, hammering the wire makes it flat, effectively reducing the cross sectional area, and increasing the length of the wire. <em>Recall that the resistivity of a metal conductor increases with length and decreases with cross sectional area. From this, we can see that the resistance of the wire will increase due to the hammering. This means Adeen is correct.</em>

For Arnell, heating a metal causes the atoms along with their electrons to vibrate in a random manner. These random motions is not organised and can be in any direction. For electricity to flow smoothly, the electrons must travel in the same direction in an orderly fashion. <em>The randomly vibrating atoms will continuously bump into the flowing electron, resisting the smooth motion of the electrons. This collisions results in an additional resistance, hence, the resistance is increased. This means Arnell is correct.</em>

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What os the term for the smallest bit into which chemical substance can be divided and still have the properties of that substan
Ostrovityanka [42]

That's a molecule of the substance.  You can break the molecule down further, into the atoms that make it up, but those don't have the properties of the original  'compound'.

Here's an example:

-- Sodium is a soft, slippery metal, that explodes when water touches it.

-- Chlorine is a poisonous green gas.

When an atom of Sodium and an atom of Chlorine combine, they make one molecule of a substance called "Sodium Chloride".  That's SALT !  It isn't green, it isn't a gas, it isn't poisonous, it isn't soft and slippery, and it doesn't explode when water touches it.

3 0
3 years ago
A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas
Arada [10]

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

7 0
3 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
What is the value of acceleration due to gravity at heght h=4re​
Norma-Jean [14]

Answer: Formula for Acceleration Due to Gravity

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Explanation:

5 0
3 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
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