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Softa [21]
3 years ago
7

An atom of arsenic has how many electron-containing orbitals

Physics
1 answer:
irina [24]3 years ago
8 0

Answer:

The Arsenic has three electron-containing orbitals. The orbitals s, p and d.

Explanation:

Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:

1s^{2}

2s^{2}

2p^{6}

3s^{2}

3p^{6}

3d^{10}

4s^{2}

4p^{3}

Therefore, the Arsenic has three electron-containing orbitals (s, p d).

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What will it be ? Plz help me
IgorC [24]

Answer:

For number 2, it will be 8.

3 0
2 years ago
Travels 11,000 feet along a dark desert highway if the car averages 84 mph find the amount of time to cover this distance
laila [671]

The time taken by traveler to cover the distance is,

t=\frac{d}{v}

Substitute the known values,

\begin{gathered} t=\frac{(11000\text{ ft)}}{(84\text{ mph)(}\frac{1.46667\text{ ft/s}}{1\text{ mph}})_{}} \\ \approx89.3\text{ s} \end{gathered}

Therefore, the time taken by traveler to cover the distance is 89.3 s.

5 0
10 months ago
During a normal reaction to a stressful event, muscles are moved to their maximum capacity, and sensitivity is
Aleonysh [2.5K]

Answer:

The paper focuses on the biology of stress and resilience and their biomarkers in humans from the system science perspective. A stressor pushes the physiological system away from its baseline state toward a lower utility state. The physiological system may return toward the original state in one attractor basin but may be shifted to a state in another, lower utility attractor basin. While some physiological changes induced by stressors may benefit health, there is often a chronic wear and tear cost due to implementing changes to enable the return of the system to its baseline state and maintain itself in the high utility baseline attractor basin following repeated perturbations. This cost, also called allostatic load, is the utility reduction associated with both a change in state and with alterations in the attractor basin that affect system responses following future perturbations. This added cost can increase the time course of the return to baseline or the likelihood of moving into a different attractor basin following a perturbation. Opposite to this is the system's resilience which influences its ability to return to the high utility attractor basin following a perturbation by increasing the likelihood and/or speed of returning to the baseline state following a stressor. This review paper is a qualitative systematic review; it covers areas most relevant for moving the stress and resilience field forward from a more quantitative and neuroscientific perspective.

Explanation:

8 0
2 years ago
A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m
belka [17]

Answer:

Energy stored in the capacitor is U=2.7\times 10^{-5}\ J        

Explanation:

It is given that,

Charge, q=1.5\ \mu C=1.5\times 10^{-6}\ C

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

U=\dfrac{1}{2}q\times V

U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36  

U = 0.000027 J

U=2.7\times 10^{-5}\ J

So, the potential energy is stored in the capacitor is U=2.7\times 10^{-5}\ J. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
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