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Tema [17]
3 years ago
10

Unlike the majority of psychologists, psychiatrists __________. A. can diagnose disorders B. use psychotherapy C. treat people w

ith disorders D. prescribe medication
Physics
2 answers:
Zigmanuir [339]3 years ago
7 0

Answer: D. prescribe medication

Psychiatrists are medical professionals or physicians who deals with psychological disorders in humans. They can prescribe medications and provide psychotherapy. Psychologists are scientists or counselled but they are not medical professionals. They can provide a psychotherapy to a patient with psychological disorder but cannot prescribe medicine like a physician.


ludmilkaskok [199]3 years ago
3 0
D prescribe medication

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Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
A tungsten wire has resistance R at 20°C. A second tungsten wire at 20°C has twice the length and half the cross-sectional area
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The resistance is 4 times the resistance of the first wire. the formula is R = p*l/A with p being resistivity, l length and A area. So if you double length and half area, which botv result in more resistance, you get p*2/0.5 or 4 (p can be abandoned because it is the same. We take standard length and area as 1)
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What is the difference between first aid and first aid kit
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Answer:

One has the word kit and the other doesn't

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Need help but the subject is science will give brainleist
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Answer:

A: decreases

B: rises

C: increases

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