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Stolb23 [73]
3 years ago
8

Research suggests that laughter improves people’s emotional and physical well-being. Write a research-based essay to inform the

reader about the positive effects of laughter on emotional and physical health. Properly cite research evidence to inform the audience about the topic. << read less stylesnormal.
SAT
1 answer:
ozzi3 years ago
8 0

It has been scientifically proven that laughter is excellent for emotional and physical well-being. The laugh can relieve physical tension, enhance the functioning of the immune system and decreases stress hormones.

It has been shown that laughter can relax facial muscles, which is fundamental for relieving physical tension.

Moreover, it has also been demonstrated that laughter can increase the heart rate and boost the adaptive immune system by generating antibodies that fight against harmful infectious agents (e.g. bacteria and viruses).

Finally, laughter can also decrease the levels of stress hormones such as cortisol, adrenaline and growth hormones.

Learn more about laughter benefits here:

brainly.com/question/14619490

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Which of the following is accurate concerning nonverbal communication.
Digiron [165]

The statement that is accurate regarding nonverbal communication is:<em> Touching the head of another person is regarded as offensive in some cultures.</em>

<h3>What is Nonverbal Communication?</h3>

Nonverbal Communication can be referred to as the mode of passing across messages to a receiver using mediums such as facial expressions, gestures, posture, body language, or eye contact.

Most cultures uses nonverbal communication, however, touching another person's head is considered offensive in some cultures.

Thus, the statement that is accurate regarding nonverbal communication is:<em> Touching the head of another person is regarded as </em><em>offensive </em><em>in some cultures.</em>

Learn more about nonverbal communication on:

brainly.com/question/5428379

7 0
3 years ago
Explore and discuss why survivors of gender based violence may feel hesitant to report this human rights violation
Naddik [55]
They feel as if they are the only ones who should deserve to questions others whiles other fee as if they deserve to question others.
6 0
2 years ago
A point charge q1 = -4. 00 nc is at the point x = 0. 60 m, y = 0. 80 m , and a second point charge q2 = +6. 00 nc is at the poin
Alekssandra [29.7K]

The net electric field is the vector sum of the components of the electric

field produced by the two charges.

The values of the magnitude and direction of the net electric field at the origin (approximate values) are;

  • 131.6 N/C
  • 12.6 ° above the negative x–axis

<h3>How are the net electric field magnitude and direction calculated?</h3>

The possible questions based on a similar question posted online are;

(a) The net electric field at the origin.

The electric field due to charge q₁ is given as follows;

\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}

Which gives;

\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C

\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C

Which gives;

\vec{E}_1 = \mathbf{21.6 \, N/C  \cdot \hat x +  28.8 \, N/C \hat y}

\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C

Therefore;

\vec  {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)

\vec  {E} = \mathbf{\left( -128.4 \, N/C  \right) \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)}

The magnitude of the net electric field is therefore;

E = \sqrt{(-128.4^2 + 28.8^2)} ≈ 131.6

  • The magnitude of the net electric field at the origin is E ≈<u> 131.6 N/C</u>

(b) The direction of the net electric field at the origin.

  • The \ direction \ is \ arctan \left(\dfrac{28.8}{-128.4} \right) \approx \underline{ 12.6^{\circ}} \ above \ the \ negative \ x-axis

Learn more about electric field strength here:

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3 0
2 years ago
A young sumo wrestler decided to go on a special high-protein diet to gain weight rapidly. He started at 90 kilograms and gained
kvv77 [185]

Answer:

90(Initial Weight) + <em>W*t = 138 Kg </em>

So the Weight increased per month is 6Kg

Explanation:

The Wrestler started at 90Kg weight

After 8 months, Wrestler had a weight of 138Kg.

The weight gained by the wrestler in '<em>t'</em> months is:

138(Kg) - 90 (Kg) = 48 (Kg)

From above equation we can see that a difference of 48Kg was attained by Wrestler.

Since the question stated that, the weight increased at a constant rate, so we can calculate the weight gained each month over a period of 8 months.

Weight gained per month: 48 (Kg) / 8 (Months) = 6 (Kg) / (Month)

So now we can write the equation in terms of <em>W</em>(Gained per month) and<em> t</em>( t number months)

The equation becomes:

90Kg (Initial Weight) + <em>W.t = 138 Kg </em>

<em>( where 'W' is Weight increased per month and 't' is number of months passed)</em>

3 0
4 years ago
Plz help ASAP plz The more resources you have to choose from during an open book test, the better you will do on the test becaus
Pani-rosa [81]
Answer: True
All the information will be in the book so that mean when you are taking the test you are able to see and find the answers
6 0
3 years ago
Read 2 more answers
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