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k0ka [10]
3 years ago
5

How do intensity, frequency, and time affect physical fitness?

Physics
1 answer:
Y_Kistochka [10]3 years ago
5 0
Important bc it makes it more effective, the specific rate makes or breaks the fitness. Frequency is important to allow your body to rebuild and repair the damage from working out, it allows the body to adapt and time for rest/ healing. Intensity depends on how much your body breaks so the recover time and frequency must be adjusted. Time effects because of the distance between frequencies which plays a role.
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How does a wind turbine transform mechanical energy into electrical energy?
Alex777 [14]

Answer:

"A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion. As wind moves past the blades of a wind turbine, it moves or rotates the blades. These blades turn a generator."

4 0
2 years ago
According to newtons third law, the two forces in a force pair act
aliina [53]
B- Same force


It’s b because force always acts in equal but opposite pairs.
6 0
3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
What are the three ways of answering a scientific question
My name is Ann [436]

Answer:

Let's start by understanding what exactly a scientific question is. A scientific question is a question that may lead to a hypothesis and help us in answering (or figuring out) the reason for some observation. A good scientific question has certain characteristics. It should have some answers (real answers), should be testable.

Here's examples of a few:

Why is that a star?

or

What is that star made of?

Hope this can lead you to the answer you're looking for at least!!

5 0
3 years ago
A beverage manufacturer wants to increase the solubility of carbon dioxide (CO2) in its carbonated drinks.
Brut [27]

D. Decreasing its temperature

Explanation:

Decreasing the temperature of the carbon dioxide gas to be dissolved in the carbonated drink will most likely increase the solubility of the gas in the drink.

Temperature has considerable effects on the solubility of gases in liquids.

  • Dissolution involves the surrounding of ions by water molecules, in this case, the carbon dioxide gas is to be surrounded by the liquid beverage medium.
  • Increasing pressure increases the rate at which gases are soluble. At high pressure, the gases are brought more in contact with the liquid medium.
  • Decreasing temperature aids gas solubility.
  • If the temperature of gases are increased,  they will not want to stay in solution as they gain a high amount of kinetic energy.
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  • Decrease in temperature and increase in pressure makes gas solubility to be fast.  

Learn more:

Rate of chemical reactions brainly.com/question/6281756

#learnwithBrainly

6 0
3 years ago
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