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3 years ago

14

How many electrons do the elements of the most stable group have in their outer energy levels

2 answers:

Wittaler [7]3 years ago

6 0

I think it might be 8 electrons

jekas [21]3 years ago

6 0

The answer is 18 electrons

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HELP ME PLEASE. i need to turn this in by tomorrow, also pls use the words above^ thank you i’ll mark u the brainliest.

Vlad [161]

Answer:

the corrdct answer is core

4 0

3 years ago

Which of the following electromagnetic waves is not used for communication?

lutik1710 [3]

Microwaves and radio waves are employed in radio and satellite communications while infrared waves are used in remote controls and infrared features of new phones and other electrons. However, gamma rays have far too much energy and cause damage to the body. They are not used in communication.
The answer is A.

8 0

3 years ago

Suppose the first coil of wire illustrated in the simulation is wound around the iron core 7 times. Suppose the second coil is w

SVETLANKA909090 [29]

Answer:

Emf induced in the coli will be equal to 26 volt

Explanation:

We have given number of turns N = 13

It is given that magnetic flux changes from 2.1 Weber to 2.7 Weber in 0.3 sec

Change in flux $d\Phi =2.7-2.1=0.6Weber$

Change in time dt = 0.3 sec

We have to find the induced emf

Induced emf is equal to $e=N\frac{d\Phi }{dt}=13\times \frac{0.6}{0.3}=26volt$

So emf induced in the coil will be equal to 6 volt

5 0

3 years ago

Read 2 more answers

For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d

Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material ( $\sigma$ ) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

Yield Stress = $\frac{Maximum load}{Area of the bar}$

⇒ $\sigma$ = $\frac{P_{max} }{A}$ ---------------- (1)

⇒ Area of the bar (A) = $\frac{\pi}{4}$ × $D^{2}$

⇒ A = $\frac{\pi}{4}$ × $45^{2}$

⇒ A = 1589.625 $mm^{2}$

Put all the values in equation (1) we get

⇒ $P_{max}$ = 200 × 1589.625

⇒ $P_{max}$ = 317925 N

In this bar the $P_{max}$ is equal to the weight of the bar.

⇒ $P_{max}$ = $M_{max}$ × g

Where $M_{max}$ is the maximum mass the bar can support.

⇒ $M_{max}$ = $\frac{P_{max} }{g}$

Put all the values in the above formula we get

⇒ $M_{max}$ = $\frac{317925}{9.81}$

⇒ $M_{max}$ = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0

4 years ago

At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t

Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

$\displaystyle{\int_{x_i}^{x_j}F(x)dx}$

and substituting with the function of our problem:

$\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}$

5 0

4 years ago