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vitfil [10]
2 years ago
9

Which is a long-term benefit of increased physical activity?

Physics
1 answer:
Veseljchak [2.6K]2 years ago
3 0

According to the research, the correct option is C. It makes bones and muscles stronger is a long-term benefit of increased physical activity.

<h3>What is physical activity?</h3>

It refers to the set of movements and actions that work the muscles, promoting energy consumption, therefore, performing physical activity frequently is essential for maintaining health.

These physical activities consume many calories and help the person not to be overweight, improving the elasticity of muscles and joints, increasing lung capacity and reducing the risk of cardiovascular disease.

Therefore, we can conclude that according to the research, the correct option is C. It makes bones and muscles stronger is a long-term benefit of increased physical activity.

Learn more about benefits of physical activity here: brainly.com/question/3478611

#SPJ1

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aksik [14]
 

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3 0
3 years ago
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
Diferencie energia cinética de energia potencial gravitacional.
Vikentia [17]

Answer:

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4 0
4 years ago
How do you know the speed of an electromagnetic wave in a vacuum?
ycow [4]
Electromagnetic waves need no matter to travel - they can travel through empty space (a vacuum). In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - which is the fastest speed possible. ...
Light traveling value through an optical Fibre is, 2 x 108 m/s. Hope that helps.
5 0
3 years ago
A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At th
Fittoniya [83]

Answer:

b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e)  T = 2π √m / k

h)   a = - A w² cos (wt+Ф)

Explanation:

a) see free body diagram in the attachment

b) We write Newton's second law

          Fe = m a

          -k x = ma

           a = -k / m x

c) the acceleration is

         a = d²x / dt²

     

      If x = A cos wt

        v = dx / dt = -A w sin (wt +Ф)

        a = d²x / dt² = - A w² cos (wt+Ф)

d) we substitute in Newton's second law

        d²x / dt² = -k / m x

   

We call

       w² = k / m

e) substitute to find w

     -A w² cos (wt+Ф) = -k / m A cos (wt+Ф)

      w² = k / m

Angular velocity and frequency are related

       w = 2π f

       f = 1 / T

       

 We substitute

      T = 2π / w

      T = 2π √m / k

g)    v= - A w sin (wt+Ф)

h) acceleration is

       a = - A w² cos (wt+Ф)

8 0
4 years ago
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