All categories

2 years ago

What component of fitness should you work on every time you exercise?

Physics

2 answers:

Pepsi [2]2 years ago

8 0

Answer:

1.body composition, (2) flexibility, (3) muscular strength, (4) muscular endurance, and (5) cardio respiratory endurance.

Explanation:

stellarik [79]2 years ago

6 0

Answer:

The component of fitness you should work on every time you exercise are your muscular strength, muscular endurance, physique (body composition), your flexibility, and endurance.

Explanation:

it is important to focus on these types of exercise in order to have a well balanced routine/fitness and to stay in fit.

hope this helps !:)

You might be interested in

Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75N/cm compressing the spring

arlik [135]

This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

$E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J$

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

$E_k = \frac{1}{2}mv^2$

And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

$E_p = 2E_k=mv^2$

From this we can determine the speed of the mass:

$E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}$

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

4 0

3 years ago

A technique in which people use machines to learn how to control their bodies is known as __________.

Debora [2.8K]

Answer:

The technique in which people use machines to learn how to control their bodies is known as D, Biofeedback.

Explanation:

Biofeedback is a variety of different machines that help people learn how to control their bodies depending on their specific needs, varying from things like scalp sensors, electrocardiographs, electromyographs and more.

7 0

3 years ago

Read 2 more answers

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago

An alpha particle (which has two protons) is sent directly toward a target nucleus with 90 protons. the alpha particle has a kin

Romashka-Z-Leto [24]

It does not move because of the sun the sun has no energy.

5 0

3 years ago

a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road

Svetradugi [14.3K]

Answer:

t = 4.1 seconds

Explanation:

It is given that,

Width of road which is to be crossed by a man is 8.25 m, it means it is distance to be covered.

Speed of man is 2.01 m/s

We need to find the time taken by the man to cross the road. It is a concept of speed. Speed of a person is given by total distance covered divided by time taken. So,

$v=\dfrac{d}{t}$

t is time taken

$t=\dfrac{d}{v}\\\\t=\dfrac{8.25}{2.01}\\\\t=4.1\ s$

So, the time taken by the man to cross the road is 4.1 seconds.

5 0

3 years ago