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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 !:)

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Explanation:

Hale-Bopp is the answer

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2 years ago

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What happens during destructive interference?

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They will subtract to form a combined wave with a lower amplitude

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2 years ago

An embryo develops inside the ________.

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2 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

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2 years ago

A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a) at 4.75 cm

b) at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula $E = \dfrac{kq }{d}$

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}$

E = 1461.95 N/C

c) The electric field E is calculated as:

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}$

E = 239.76 N/C

7 0

2 years ago