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3 years ago

15

Which type of stress is most likely the cause of irritability?

1 answer:

masha68 [24]3 years ago

4 0

Answer:

The correct answer is - distress.

Explanation:

Eustress is a type of stress that is beneficial to person and caused by adrenaline. Hypostress is a stress that is caused due to boredom or no work. It is stress comes with wasting the time.

Distress is the bad stress caused by anxiety, sadness, grief, and such type of stress that is caused irritability. Irritability is caused not just by stress but many factors.

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Whose geocentric model was accepted for 1400 years

zysi [14]

<span>IDK the answer but I think it was "Ptolemy" whose geocentric model of the solar system was accepted for 1,400 years, and was embraced by the the Church until Galileo called it into question.</span>

8 0

4 years ago

Monochromatic light from a point source is directed toward the sharp edge of a solid object. How does diffraction change the ima

Anna71 [15]

Did you get the answers the quiz??? If so please write out the answers in the comment section :)

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4 years ago

Read 2 more answers

A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha

USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

$\Phi_E=\frac{Q}{\epsilon_o}$

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

$\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}$

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

$Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C$

Finally, you obtain for E:

$E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}$

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0

3 years ago

How many items are present in the compound (NH4)2Cr2O7

SVEN [57.7K]

In 2Cr2O7 there’s 2 items; Cr and O

5 0

3 years ago

A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C

Rus_ich [418]

Answer:

$1.50\ *10^{-6} }$

Explanation:

Given

e=100 N/C

M=0.15 g

$q=\ 22\ pC\\=\ 22\ *10^{-2}$

The ratio of the electric force on the bee to the bee's weight can be determined by the following formula

$\frac{fe}{M*9.81}$

$\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}$

$=\ 1.50\ *10^{-6}$

4 0

4 years ago