All categories

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.

You might be interested in

Which statement best describes beaches? Beaches change over time as waves move further up the shore. Beaches are always changing

NemiM [27]

Answer:

I think the 1st statement is right.

Explanation:

Wind patterns doesn't stay the same.

Waves don't follow the same patterns.

Waves move further up the shore.

I didn't hear about "waves adding" before..so i guess 1st statement is right.

3 0

3 years ago

Read the passage below and answer the question.

yuradex [85]

Answer: ?

Explanation:

4 0

3 years ago

Read 2 more answers

Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and obser

Andrej [43]

Answer:

Answer is C. Both technicians A and B.

Refer below.

Explanation:

Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and observed to see if the vacuum drops with the engine at 2500 RPM for 30 seconds. Technician B says that a pressure gauge can be used to check for backpressure. The following technician is correct:

Both technicians A and B

7 0

3 years ago

Read 2 more answers

How do you add this vector graphically

arlik [135]

(-3,3)+(2,3)=(1,6)

this is the answer :)

8 0

3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

6 0

3 years ago