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3 years ago

Important things about emotional health

Physics

2 answers:

anastassius [24]3 years ago

7 0

Having a good mental health (emotional health) gives you these following benefits:

Decreases:

-Anger

-Anxiety (worry and fear)

-Confusion

-Depression (you’ll likely be better at preventing, reducing, and managing depression)

-Headaches

-Stress and tension (you’ll likely be able to cope better with stress)

Increases:

+Assertiveness (being able to ask for what you need and make decisions)

+Confidence and feeling able to do things

+Emotional stability (less troubled by life’s challenges and disappointments)

+Independence

+Memory

+Having a positive mood

+Perception (better at noticing what’s going on around you)

+Positive body image (feeling good about the way you look)

+Feeling of well-being

+Self-worth and self-esteem (feeling good about the way you see yourself)

Hope this helps.

atroni [7]3 years ago

4 0

Answer:

The important thing about emotional health is being in control of their thoughts, feelings, and also behaviors.

Explanation:

Emotional health is an important part of overall health. People who are emotionally healthy are in control of their thoughts, feelings, and behaviors. They're able to cope with life's challenges. They can keep problems in perspective and bounce back from setbacks.

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What is the weight of a dog that has a mass of 47.0 kg?

Nat2105 [25]

Answer:

while we often confuse mass with weight, 47 kg is 47 x 9.8 = 460.6 Newtons. 9.8 is acceleration of gravity in m/sec/sec

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3 years ago

Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 m

Savatey [412]

Answer:

b) 0.5 N

Explanation:

From coulomb's law,

F = kq'q/r².................... Equation 1

Where F =force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

q'q = Fr²/k........................... Equation 2

Given: F = 2 N, r = 1 m, k = 9.0×10⁹ Nm²/C²

Substituting into equation 2

q'q = 2(1)²/(9.0×10⁹)

q'q = 2/9.0×10⁹ C².

If the distance between the charges is increased to 2 meters,

r = 2 m, q'q = 2/9.0×10⁹ C².

Substitute into equation 1

F = 9.0×10⁹(2/9.0×10⁹)/2²

F = 2/4

F = 1/2 = 0.5 N.

The right option is b) 0.5 N

5 0

4 years ago

g A has all the mass at the rim, while wheel B has the mass uniformly distributed, like a solid disk. The wheels have the same m

marishachu [46]

Answer:

The second wheel

Explanation:

The torque is given by

$\tau=I\alpha$ (1)

where I is the moment of inertia and a is the angular acceleration. If we take into account the moment of inertia of a disk and a ring ()for the first wheel) we have:

$I_r=mR^2\\I_d=\frac{1}{2}mR^2$

where we used that both wheel have the same mass. By replacing in (1) we obtain:

$\alpha_r=\frac{\tau}{I_r}=\frac{\tau}{mR^2}\\\alpha_d=\frac{\tau}{I_d}=\frac{\tau}{\frac{1}{2}mR^2}=2\alpha_r\\$

Hence, the second wheel (the disk) has a greater acceleration.

hope this help!!

5 0

4 years ago

Read 2 more answers

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

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3 years ago

An object has a mass of 8.00 kg. what is the gravitational force of the object on earth

KonstantinChe [14]

When a mass of 8 kg is located on the Earth's surface, the magnitude of each of
the gravitational forces attracting the mass and the Earth toward each other is

   (mass) x (acceleration of gravity on Earth) =

      (8.0 kg) x (9.81 m/s²) =

      78.48 kg-m/s² = <em>78.48 newtons </em>(about 17 pounds 10.2 ounces)

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4 years ago

Read 2 more answers