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babunello [35]
2 years ago
13

Why are slow twitch muscles more beneficiall than fast twitch mucles for cardiorespiratory fitness

Physics
2 answers:
Dmitriy789 [7]2 years ago
4 0

Answer:

otherwise known as option D

Explanation:

:)

Lady_Fox [76]2 years ago
4 0

Answer:

D Slow-twitch muscles are able to use oxygen more efficiently than fast-twitch muscles.

is the answer

Explanation:

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When jogging outside you accidently bump into a curb. Your feet stop but your body continues to move forward and you end up on t
oee [108]
Inertia I think because I've heard it around school and in science
6 0
3 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}  = \frac{725}{\sqrt{725}}  = \sqrt{725} ft / sec

5 0
3 years ago
Since the coefficient of friction is less than 1, what does that mean about the normal force and the force of friction?
ludmilkaskok [199]

Answer:

there is friction between the two things

Explanation:

7 0
2 years ago
A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from
Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0
2 years ago
Read 2 more answers
During an experiment, you take a measurement of 12.9 inches.
svlad2 [7]

Answer:

1 inch = 2.54 cm

12.9 inches= 12.9 x 2.54

= 32.766

= 32.8 cm (approximately)

Hope it helps...

7 0
3 years ago
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