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3 years ago

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

Physics

2 answers:

Pepsi [2]3 years ago

8 0

Answer:

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

Explanation:

stellarik [79]3 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 !:)

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Which of the following is not a tectonic activity?

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Answer:

Erosion of Mountains

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3 years ago

When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s . By how much has thermal energy increased duri

Advocard [28]

Complete question:

A fireman of mass 80 kg slides down a pole. When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s. By how much has thermal energy increased during his slide?

Answer:

The thermal energy increased by 3,099.2 J

Explanation:

Given;

mass of the fireman, m = 80 kg

initial position of the fireman, hi = 4.2 m

final speed, v = 2.2 m/s

The change in the thermal energy is calculated as;

ΔE + (K.Ef - K.Ei) + (Uf - Ui) = 0

where;

ΔE is the change in the thermal energy

K.Ef is the final kinetic energy

K.Ei is the initial kinetic energy

Uf is the final potential energy

Ui is the initial potential energy

$\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i)=0\\\\initial \ velocity, \ v_i = 0\\final \ height , \ h_f = o\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2) + ( - mgh_i)=0\\\\\Delta E_{th} + \frac{1}{2} mv_f^2 - mgh_i = 0\\\\\Delta E_{th} =mgh_i - \frac{1}{2} mv_f^2\\\\\Delta E_{th} = 80 \times 9.8 \times 4.2 \ \ - \ \ \frac{1}{2} \times 80 \times (2.2)^2 \\\\\Delta E_{th} = 3292.8 \ J \ - \ 193.6 \ J\\\\\Delta E_{th} = 3,099.2 \ J$

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3 years ago

Plzz helpppp meee!!!!

spin [16.1K]

A=f/m
A=900/425
A=2.18

To determine acceleration you divide the force by the mass.

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A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicula

melomori [17]

Answer:

Explanation:

Moment of inertia of the rod = 1/12 m L²

m is mass of the rod and L is its length

= 1/2 x 2.3 x 2 x 2

= 4.6 kg m²

Moment of inertia of masses attached with the rod

= m₁ d² + m₂ d²

m₁ and m₂ are masses attached , and d is their distance from the axis of rotation

= 5.3 x 1² + 3.5 x 1²

= 8.8 kg m²

Total moment of inertia = 13.4 kg m²

B )

Rotational kinetic energy = 1/2 I ω²

I is total moment of inertia and ω is angular velocity

= .5 x 13.4 x 2²

= 26.8 J .

C )

when mass of rod is negligible , moment of inertia will be due to masses only

Total moment of inertia of masses

= 8.8 kg m²

D )

kinetic energy of the system

= .5 x 8.8 x 2²

= 17.6 J .

3 0

3 years ago