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3 years ago

What species has the electron configuration [ar]3d2?

Physics

2 answers:

Brrunno [24]3 years ago

8 0

The given configuration is : [Ar]3d²

The inner shells of this species has the noble gas, Argon (Ar) configuration while its valence configuration is 3d².

Now the atomic number of Argon = 18

Therefore, the species will have an atomic number = 18 + 2 = 20

Based on the periodic table the species with atomic number 20 is calcium

Ca = [Ar] 3d²

soldier1979 [14.2K]3 years ago

7 0

Answer:

Explanation:

The electronic configuration of the Argon (Ar) has atomic number 18 with all its layers filled

To fill the 3d2 level, you must first fill the 4s level with its two electrons, since it has less energy so the electronic configuration of the atom would be

[Ar] 4s² 3d² corresponding to Titanium

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In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g

Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η= $\frac{electric power}{water power}$

Plugging the values in the above equation

0.84=EP/57.38

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Therefore, the electric output of the plant is 48.19 MW.

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2 years ago

Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile? Answ

Vinil7 [7]

Answer: 272.82 drop/tile

Explanation:

Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?

Tiles/ft^2 × drop/tiles = drop/ft^2

Tiles will cancel out. Leaving the answer to be drop/ ft^2

Substitutes all the magnitude of the above units.

17 × drop/tiles = 4638

Make drop/tiles the subject of formula

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Therefore, 272.82 drop/tile drops fall on each tile?

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3 years ago

A bike with 15cm diameter wheels accelerates uniformly from rest to a speed of 7.1m/s over a distance of 35.4m. Determine the an

Finger [1]

Answer:

9.47 rad/s^2

Explanation:

Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,

s = 35.4 m

let a be the linear acceleration.

Use III equation of motion.

v^2 = u^2 + 2 a s

7.1 x 7.1 = 0 + 2 x a x 35.4

a = 0.71 m/s^2

Now the relation between linear acceleration and angular acceleration is

a = r x α

where, α is angular acceleration

α = 0.71 / 0.075 = 9.47 rad/s^2

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3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

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3 years ago

Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea

notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

$v=u+at$

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

$0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s$

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

3 0

3 years ago

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