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4 years ago

8

The person given credit for the first modern atomic theory was __________.

1 answer:

mote1985 [20]4 years ago

7 0

<span>The person given credit for the first modern atomic theory was __________.

Dalton.</span>

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PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat

Svetach [21]

Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of 1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx

Answer: 20 + 3421 = 3441 °C

4 0

3 years ago

What is the difference between the number of electrons in an atom of selenium, Se, and the number of electrons in an atom of alu

Oxana [17]

Well, electrons can be converted into a atomic number so if SE atomic number is 34 that means it has 34 electrons. AI has a atomic number of 13 meaning it has 13 electrons.

So the difference is that SE has more electrons then AI.

Hope this helped. :D

4 0

3 years ago

Read 2 more answers

When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea

crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = $\frac{(172\ \mu C)}{V}$

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = $\frac{(172+220)\ \mu C)}{V}$

or

C' = $\frac{(392)\ \mu C)}{V}$

now the dielectric constant (k) is given as:

$k=\frac{C'}{C}$

substituting the values, we get

$k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}$

or

k = 2.279

6 0

3 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

4 years ago

A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

nadya68 [22]

Answer:

Explanation:

Given

mass of steel ball $m=0.2\ kg$

initial speed of ball $u=10\ m/s$

Final speed of ball $v=-10\ m/s$ (in upward direction)

Impulse imparted is given by change in the momentum of object

therefore impulse J is given by

$J=\Delta P$

$\Delta P=m(v-u)$

$\Delta P=0.2(-10-10)$

$\Delta =-4\ N-s$

so magnitude of Impulse =4 N-s

7 0

3 years ago

Read 2 more answers