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Ira Lisetskai [31]

3 years ago

12

what are valence shell electrons?a. electrons thata re unchargedb. electrons that are positively chargedc. electrons that are av

ailable for bondingd. electrons that are attached to the nucleeus

1 answer:

goblinko [34]3 years ago

6 0

Your answer is c. electrons that are available for bonding.

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What element exists in almost every compound in your body and other living things

Aleks [24]

Carbon is one of the main building blocks of life. This is what carbon dating is so effective, because scientists are able to tell the approximate age of something that was once alive given how much carbon is still in the animal.
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3 0

3 years ago

Your roommate leaves a 120W fan running in your apartment.Over the course of an hour,how much thermal energy does the fan add to

zubka84 [21]

Answer:

$4.32\cdot 10^5 J$

Explanation:

Power is related to energy by the following relationship:

$P=\frac{E}{t}$

where

P is the power used

E is the energy used

t is the time elapsed

In this problem, we know that

- the power of the fan is P = 120 W

- the fan has been running for one hour, which corresponds to a time of

$t = 1 h \cdot (60 min/h)(60 s/min)=3600 s$

So we can re-arrange the previous equation to find E, the energy (in the form of thermal energy) released by the fan:

$E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J$

3 0

3 years ago

Read 2 more answers

Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it

jek_recluse [69]

Answer:

a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

I = I₁ cos² θ'

we substitute

I = (I₀ cos² tea) cos² (θ - 90)

cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

I = Io cos² θ sin² θ

1= cos²θ+ sin²θ

sin²θ = 1 - cos²θ

I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

$\frac{dI}{d \theta}$ = 0

\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

cos θ - 2 cos³ θ = 0

cos θ ( 1 - 2 cos² θ) = 0

The zeros of this function are in

θ = 90º

1-2cos²θ =0 cos θ = 0.25 θ = 75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0

2 years ago

11. Explain why in terms of gravity and air resistance and the 2nd law, objects in free fall regardless of mass hit at the same

Alchen [17]

Answer:

When an object is dropped, it accelerates toward the center of Earth. Newton's second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force.

7 0

3 years ago

Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

Harman [31]

Answer: a) 11.76 m/s b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

$V_{f}=V_{o}+at$ (1)

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (2)

Where:

$V_{f}$ Is the final velocity of the object

$V_{o}=0$ Is the initial velocity of the object (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d$ is the height of the tower

$t=0.02min=1.2 s$ is the time it takes to the object to reach the ground

b) Begining with (1):

$V_{f}=0+at$ (3)

$V_{f}=at=(9.8 m/s^{2})(1.2 s)$ (4)

$V_{f}=11.76 m/s$ (5) This is the final velocity of the object

a) Substituting (5) in (2):

$(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d$ (6)

Clearing $d$ :

$d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})}$ (7)

$d=7.056 m$ (8) This is the height of the tower

4 0

3 years ago