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Nuetrik [128]
3 years ago
14

Which of the following BEST describes cardiovascular fitness?

Physics
2 answers:
nikdorinn [45]3 years ago
8 0

Hi! In order for me to help you, I need the answer choices. Thanks! -AmbitiousAndProud

AnnyKZ [126]3 years ago
3 0

Answer:the answer is A.

Explanation:

Just got 100%

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The sun's energy begins as what form of energy?
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nuclear energy.............

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The correct field line would be A.
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Monica [59]

Answer:

im pretty sure it is C. Insulation

Explanation:

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5 0
3 years ago
An airplane weighing 11,000 N climbs to a
Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

W=(mg)\Delta h

where

mg = 11,000 N is the weight of the airplane

\Delta h = 1.6 km = 1600 m is the change in height

Substituting,

W=(11,000)(1600)=17.6\cdot 10^6 J

Now we can calculate the power delivered, which is given by

P=\frac{W}{t}

where

W=17.6\cdot 10^6 J is the work done

t=9.8 min \cdot 60 = 588 s is the time taken

Substituting,

P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W

Finally, we can convert the power into horsepower (hp), keeping in mind that

1 hp = 746 W

Therefore,

P=\frac{2.99\cdot 10^4}{746}=40.1 hp

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
3 years ago
A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and
icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

  • g = acceleration due to gravity = 9.8 m/s²,
  • r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
  • ρ = density of sphere = 19 g/cm³,
  • σ = density of water = 1.0 g/cm³ and
  • η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

brainly.com/question/21684177

#SPJ1

4 0
1 year ago
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