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3 years ago

List 5 plants or animals that the emporer penguin eats

Physics

1 answer:

patriot [66]3 years ago

6 0

Answer:

Emperor penguins are near the top of the Southern Ocean’s food chain. They have a varied menu that changes with the season. Some prey items are more important than others. One of the most frequently eaten prey species is the Antarctic silverfish Pleuragramma antarcticum. They also eat other fish, Antarctic krill and some species of squid. Most prey items are small. Since they are very cold when ingested, their small size makes it easier to bring food up to body temperature to digest it.

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Find the work done by the force field F(x, y) = xi + (y + 6)j in moving an object along an arch of the cycloid r(t) = (t − sin(t

Feliz [49]

Answer:

177.65

Explanation:

Work done by the force field,F along path C is given by:

$W=\int\limits_C F.dr$

Given that:

$r(t) = (t -sin(t)i + (1 -cos(t)j,\\\\\frac{dr}{dt}=(1-cos t)i+sin t\ j\\\\dr=(1-cos\ t)i +sin\ t \ j)dt\\\\F(x,y)=x\ i +(y+6)j\\\\F(r(t))=(t-sin \ t) i+((1-cos \t)+2)j\\\\\\$

$F(r(t))=(t-sin \ t)i+(3-cos \ t)j\\\\\\W=\int\limits_C F.dr\\=\int\limits^{6\pi}_0(t-sin \ t)i+(3-cos \ t)j).((1-cos \ t)i+sin \ t \ j)dt\\\\=\int\limits^{6\pi}_0(t-sin \ t)(1-cos \ t)+(3-cos \ t)sin \ t \ dt\\\\=\int\limits^{6\pi}_0t-tcos \ t+2sin\ t\ dt\\\\=\int\limits^{6\pi}_0-tcos \ t\ dt+[\frac{t^2}{2}-2cos \ t]\limits^{6\pi}_0\\\\=-I+177.65$

#Integrating I by parts:

$I=\int\limits^{6\pi}_0 tcos \ t \ dt\\\\=[\int \ tcos \ t \ dt]\limits^{6\pi}_0\\\\=[t\int cos \ t \ dt-\int (\frac{dt}{dt}\intcos \ t \ dt)dt]\limits^{6\pi}_0\\\\=[tsin\ t -\int sin\ t \ dt]\limits^{6\pi}_0\\\\=0$

$W=0+177.65$

Hence, work done is 177.65

3 0

3 years ago

When a +0.00235 C charge

irakobra [83]

Answer:177.87

Explanation:

4 0

3 years ago

Increase the mass of the beanbag its gravitational potential energy will

kotegsom [21]

The gravitational potential energy will increase

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in a gravitational field.

Near the Earth's surface, the GPE of an object is given by

$GPE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object above the ground

From the equation, we see that the GPE is directly proportional to the mass: therefore, if the mass increases, the GPE will increase as well.

So, for the beanbag in this problem, when its mass increases, the GPE will increase as well.

Learn more about gravitational potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

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5 0

3 years ago

Salmon often jump waterfalls to reach their breeding grounds starting downstream, 2.9 meters away from a waterfall .436 meters i

nikklg [1K]

The minimum speed must a Salmon jumping with to leave the water

to continue upstream is 5.79 m/s

Explanation:

At first let us find the two component of the jumping velocity of the fish

1. Horizontal component $u_{x}$ = u cosФ

2. Vertical component $u_{y}$ = u sinФ

where u is the initial velocity and Ф is the angle between the horizontal

and the initial velocity u

→ Ф = 44.7°

→ $u_{x}$ = u cos(44.7)

→ $u_{y}$ = u sin(44.7)

The horizontal distance x is 2.9 meters away from a waterfall

The vertical distance y is 0.436 meters

3. The horizontal distance x = $u_{x}$ t

4. The vertical distance y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity

→ x = u cos(44.7) t

→ x = 2.9 meters

→ 2.9 = u cos(44.7) t

Divided both sides by u cos(44.7)

→ t = $\frac{2.9}{ucos(44.7)}$ ⇒ (1)

→ y = u sin(44.7) t + $\frac{1}{2}$ gt²

→ y = 0.436 meters , g = -9.81 m/s²

→ 0.436 = u sin(44.7) t - 4.905 t² ⇒ (t)

Substitute (1) in (2) to make the equation of u only

→ 0.436 = u sin(44.7)( $\frac{2.9}{ucos(44.7)}$ ) - 4.905 ( $\frac{2.9}{ucos(44.7)}$ )²

→ 0.436 = 2.9 ( $\frac{sin(44.7)}{cos(44.7)}$ - $\frac{41.25105}{u^{2}[cos(44.7)]^{2}}$

→ 0.436 = 2.8698 - $\frac{81.4671}{u^{2} }$

Subtract 2.8698 from both sides

→ -2.4338 = - $\frac{81.4671}{u^{2} }$

Multiply both sides by -1

→ 2.4338 = $\frac{81.4671}{u^{2} }$

By using cross multiplication

∴ 2.4338 u² = 81.4671

Divide both sides by 2.4338

→ u² = 33.4732

Take √ for both sides

→ u = 5.79 m/s

<em>The minimum speed must a Salmon jumping with to leave the water</em>

<em>to continue upstream is 5.79 m/s </em>

Learn more:

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4 0

3 years ago

What does intonation refer to select one of the options below as your answer:a. the change in the pitch (high or low sound of th

Kaylis [27]

I associate "Tone" with sound frequencies. Tonic solfa being a music example which is basically, I think, using a certain musical key signature such as C major. In a major scale the sequence of "musical intervals" (ie frequency steps) is key note TONE TONE SEMITONE ... TONE TONE SEMITONE.octave higher than keynote.
So, I'd go for a here.

3 0

3 years ago