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Aleksandr [31]
2 years ago
5

Exercise provides a healthy outlet for feelings, which helps improve __________. A. emotional health B. mental health C. spiritu

al health D. environmental health
Physics
2 answers:
miss Akunina [59]2 years ago
6 0

Answer: A. Emotional health

Exercise provides a healthy outlet for feelings, in this way the exercise improves the feelings. Regular exercise can have a positive impact on the symptoms of anxiety, depression, relieves stress, it improves memory, it can help in sleeping better and also boosts the mood of the individual. Therefore, it can be said that exercise helps in improving the overall emotional health.

Juliette [100K]2 years ago
5 0

Answer: Option (A) is the correct answer.

Explanation:

Emotions are the output of feelings. When we exercise, we tend to feel healthy in terms of emotions, mental well being, spiritual well being etc.

As a result, exercise bring a lot of change in our health and attitude. But when exercise provides a healthy outlet for feelings, then it helps improve emotional health.

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A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an
Nadya [2.5K]
Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is 
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
   = (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
   = 4.5 cm

Answer: 4.5 cm
7 0
2 years ago
Read 2 more answers
Kate is researching air pollution and finds some information on ozone. She knows that ozone is a good thing as part of the ozone
jek_recluse [69]

"At ground level, ozone contributes to smog" so it is also an air pollutant.

Option: A

<u>Explanation</u>:

ozone is naturally present in stratosphere and acts as shield against harmful ultraviolet radiations. But it acts a pollutant contributing to global warming when it is present in lower level atmosphere particularly troposphere. In this level it combines with primary pollutants that is "nitrogen oxides" and "volatile organic" compounds to form secondary pollutant which absorbs outgoing radiation and contributes in raising the temperature. It has harmful impacts on vegetation as well as human health.

7 0
2 years ago
The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u
Aleksandr [31]
Let
upthrust = T
weight = W = mg
Air resistance = F

When balloon is descending, air resistance acts upwards (positive)
By Newton's first law, the net force on the balloon is zero, or
T+F-W=0......................(1)

Let w=weight of material dumped so that balloon now travels upwards at constant speed.
Air resistance acts against motion, namely downwards.
The Newton's equation now reads
T-F-(W-w)=0................(2)

Subtract (2) from (1)
T+F-W - (T-F-(W-w)) = 0
Solve for w
w=2F, or
the WEIGHT of material to be released equals twice the resistance of air.
3 0
2 years ago
A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric
Tcecarenko [31]

Answer:

a)   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m  and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s   c) v_{z}  = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

         Fe + Fm = m a

         a = (Fe + Fm) / m

the electric force is

         Fe = q E   k ^

         Fe = 4.25 10-4 60 k ^

         Fe = 2.55 10-2 k ^

the magnetic force is

         Fm = q v x B

         Fm = 4.25 10⁻⁴  \left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]

         fm = 4.25 10⁻⁴ (-j ^ 30 4)

         fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

      aₓ = 0

there is no force

Axis y

      ay = -5.10 10²/5 10⁻⁵ j ^

      ay = -1.02 107 j ^ / s2

z axis

      az = 2.55 10⁻² / 5 10⁻⁵ k ^

      az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

           x = vo t + ½ aₓ t₂2

           x = 30 0.02 + 0

           x = 0.6m

       

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

           y = I + go t + ½ ay t²

           y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

           y = 1 - 2.04 10³

           y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

          z = zo + v₀ t + ½ az t²

          z = 0 + 0 + ½ 5.1 10² 0.02²

          z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

 since there is no acceleration the speed remains constant

          vₓ = 30.0 m / s

Axis y

  let's use the equation v = v₀ + a_{y} t

         v_{y} = 0 + -1.02 10⁷ 0.02

          v_{y} = 2.04 10⁵ m / s

z axis

          v_{z} = vo + az t

          v_{z} = 0 + 5.1 10² 0.02

          v_{z}  = 1.02 10⁻¹m / s

8 0
2 years ago
A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m
Marrrta [24]

Answer:

-1.47 m/s^2

Explanation:

We can use the following SUVAT equation to solve the problem:

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2

4 0
2 years ago
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