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bagirrra123 [75]

3 years ago

11

What was the significant change that Glenn Seaborg made to Mendeleev’s periodic table?

1 answer:

stira [4]3 years ago

7 0

Answer:

The Seaborg introduced the whole new row to the periodic table called actinides.

Explanation:

The american scientist Glenn Seaborg has many important contribution the chemistry. His work made a significant change in the mendeleev's periodic table. He introduced new row that is placed below the lanthanides row, called actinides.

He was co-discoverer or principle discoverer of plutonium and einsteinium. He was responsible for the discovery of more than 100 medical isotopes.

Thus, the Seaborg introduced the whole new row to the periodic table called actinides.

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A 1-megabit computer memory chip contains many 27 fF capacitors. Each capacitor has a plate area of 3.09 × 10−11 m 2 . Determine

valentina_108 [34]

Answer:

Plate separation of each capacitor is 101.132 °A

Explanation:

The formula to calculate the capacitance in empty space as a function of distance (square parallel plates) is:

$C=\epsilon_{0}\frac{Area}{distance}$

clearing for distance:

$distance=\epsilon_0 \frac{Area}{Capacitance} \\\\\epsilon_0=8.8542(10)^{-12}C^2/Nm\\\\Area=3.09(10)^{-11}m^2\\Capacitance=27(10)^{-15}F\\\\Replacing\\\\distance=\frac{8.8542(10)^{-12}*3.09(10)^{-11}}{27(10)^{-15}} =1.0133(10)^{-8}m\\\\In A\\distance= 101.132 A$

8 0

3 years ago

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

5 0

3 years ago

Pressure __________ with depth to support the fluid weight above

eduard

Answer:Hydrostatic

Explanation: I think this is the answer, not sure. Sorry

5 0

3 years ago

g An inductor used in a dc power supply has an inductance of 12.0 H and a resistance of It carries a current of 0.300 A. (a) Wha

valkas [14]

Answer:

Explanation:

Energy of an inductor = 1/2 L i²

L is inductance , i is current .

= 1/2 x 12 x .3²

= .54 J

4 0

3 years ago

Dave and Alex push on opposite ends of a car that has a mass of 875 kg. Dave pushes the car to the right with a force of 250 N,

White raven [17]

The net force acting on the car is 65 N to the left

The net force acting on an object is simply defined as the resultant force acting on the object.

From the question given, we obtained the following data:

Force applied to the right (Fᵣ) = 250 N
Force applied to the left (Fₗ) = 315 N
Net force (Fₙ) =?

The net force acting on the car can be obtained as follow:

Fₙ = Fₗ – Fᵣ

Fₙ = 315 – 250

<h3>Fₙ = 65 N to the left </h3>

Therefore, the net force acting on the car is 65 N to the left

Learn more on net force: brainly.com/question/19549734

8 0

2 years ago

Read 2 more answers