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marin [14]
3 years ago
11

An airplane weighing 11,000 N climbs to a

Physics
1 answer:
Gennadij [26K]3 years ago
8 0

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

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A spring with a spring constant k of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. It is then st
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Answer:

Explanation:

angular frequency of resulting SHM ω = √(k/m

ω  =√(100x32/1

ω  = 56.57

Period T

ω = 2π /T = 56.57

T = 2π / 56.57

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A = 1 inch

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x ( t ) in inch  =  A cosω t

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2. Two vectors with magnitudes of 1 meters and 3 meters cannot have a resultant of (5.0 points)
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The intensity of a sunspot is found to be 3 times smaller than the intensity emitted by the solar surface. What is the approxima
Levart [38]

Answer:

If the temperature of the solar surface is 5800 K then the approximate temperature of the sunspot is a) 4400 K.

Explanation:

The most straightforward way to solve this is using Stefan-Boltzmann law that states that I the energy radiated per unit surface area per unit time (watt per unit area \frac{W}{m^{2}}) of a black body is proportional to the fourth power of the temperature T of the body:

                                            I=\sigma T^{4}

with \sigma=5.67x10^{-8} Wm^{-2} K^{-4} being the Stefan constant.

A black body is an idealized physical body that is a perfect absorber because it absorbs all incident electromagnetic radiation and is also an ideal emitter. The Sun is considered to be a black body at different layers and different temperatures.

We are told that the intensity of a sunspot I_{sunspot} is found to be 3 times smaller than the intensity emitted by the solar surface I_{surface}, that means that:

                                   I_{sunspot}=\frac{I_{surface}}{3}

then using the expression of Stefan-Boltzmann law we get that

                                   \sigma T_{sunspot} ^{4}=\sigma T_{surface} ^{4}

we cross out \sigma and use the fourth root in each side of the equation

                                   \sqrt[4]{T_{sunspot} ^{4}}=\frac{\sqrt[4]{T_{surface} ^{4}}}{\sqrt[4]{3}}

                                   T_{sunspot}=\frac{T_{surface}}{\sqrt[4]{3} }

then we use that

  • T_{surface}=5800 K
  • \sqrt[4]{3}\approx1,316

                                 

                                   T_{sunspot}=\frac{5800 K}{1,316}

                                   T_{sunspot}=4407,3 K

So finally we get that

                                   T_{sunspot}\approx4400 K

6 0
3 years ago
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