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iris [78.8K]
3 years ago
5

Which structure found in echinoderms can sometimes perform a similar function as a radula found in some mollusks? This Is A Scie

nce Question
A.foot
B.gill
C.mantle
D.stomach
Physics
2 answers:
kow [346]3 years ago
8 0

Answer: Foot.

Explanation:

Echinoderm means spiny skin. Some example of echinoderms are star fish, sea urchin, sand dollar.

These organisms have no eyes, no brains but they posses tube feet. This feet helps them in respiration, feeding and locomotion.

Tube feet helps in passing food to the oral cavity. This organ performs the similar function to radula in mollusks.

hammer [34]3 years ago
6 0
I think it’s D which is the stomach if not I can get you someone that can help you
You might be interested in
A runner starts from rest, runs for 30
aleksklad [387]

Answer:

u=0

t=30s

a=2m/s²

v=?

v= u+at

= 0+(30)(2)

= 60m/s

6 0
3 years ago
How long does it take a P-wave to travel 7,000 km? ______ minutes b) How long does it take an S-wave to travel 7,000 km? ______
lukranit [14]

Answer:

A. 8.64 secs.

B. 14.58 secs.

C. 26.002 secs.

D. 33.46secs.

Explanation:

A. P wave would travel 7000km

p-wave travels on a speed of 13.5km/s

= 7000km/13.5km/s

= 8.64 secs.

B. S-wave time to travel 7000km

s-wave travels on a speed of 8km/s

= 7000km/8km/s

= 14.58 secs.

C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).

= 7000km to miles

= 4349.598m/2.788m/s

= 26.002 secs.

D. Rayleigh wave to travel 7,000 km

10,000m/s ( 2.1667 m/s ).

= 7000km to miles

= 4349.598m/2.1667m/s

= 33.46secs.

5 0
4 years ago
Suppose you have a cylinder filled with diatomic oxygen (O2) and it is running low. The cylinder is shown above, is made of stee
Rudiy27

Answer:

The number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.

Explanation:

The number of oxygen molecules can be found using the Ideal Gas law:

PV = nRT        

Where:

P: is the pressure = 100 psi

V: is the volume = 10 L

n: is the number of moles =?

T: is the temperature = 20 °C = 293 K

R: is the gas constant = 0.082 L*atm/(K*mol)

Hence, the number of moles is:

n = \frac{PV}{RT} = \frac{100 psi*\frac{1 atm}{14.7 psi}*10 L}{0.082 L*atm/(K*mol)*293 K} = 2.83 moles

Now, the number of molecules can be found with Avogadro's number:

n_{m} = \frac{6.022 \cdot 10^{23}\: molecules}{1\: mol}*2.83 moles = 1.70 \cdot 10^{24} \: molecules

Therefore, the number of O₂ molecules that are left in the cylinder is 1.70x10²⁴.

I hope it helps you!              

3 0
3 years ago
greg has a hoop and a solid cylinder, and wants to spin each around its axis of rotation. they have the same mass and radius. wh
brilliants [131]

Both moments of inertia solid cylinder and hoop are the same.

We need to know the rotational inertia of a solid cylinder and hoop.

Solid cylinder inertia is given by

I = 1/2 x M x R²

Hoop cylinder inertia is given by

I = 1/2 x M x (R1 + R2)²

where I is a moment of inertia, M is mass, R is the total radius, R1 is central radius and R2 is skin radius.

The total radius can be defined by

R = R1 + R2

hence R² is

R = (R1 + R2)²

Substitute to solid cylinder inertia

I = 1/2 x M x R²

I = 1/2 x M x (R1 + R2)²

Hence, both moments of inertia solid cylinder and hoop are the same.

Find more on moment of inertia at: brainly.com/question/14460640

#SPJ4

6 0
1 year ago
Can someone explain with the Answers pls?
Dovator [93]

Answer:

a) The frequency of the third harmonic is 786 Hz

b) The frequency of the first harmonic is 340 Hz

c) The frequency of the fifth harmonic is 1640 Hz

Explanation:

The rule is as follows:

If the first harmonic frequency (also called the fundamental frequency) is F, then:

The frequency of the second harmonic (also called the second overtone) is:

2*F

The frequency of the third harmonic (also called the third overtone) is:

3*F

And so on.

With this information, we can answer the questions:

a) We want to find the frequency of the third harmonic, such that the frequency of the first harmonic is 262 Hz.

Then we have F = 262Hz

And the frequency of the third harmonic will be:

3*F = 3*262Hz = 786 Hz

b) First harmonic for a string whose fifth harmonic frequency is 1700Hz.

Let's define F as the first frequency (the one we want to find)

Then the fifth harmonic frequency can be written as:

5*F = 1700Hz

With this equation we can find the value of F:

F = 1700Hz/5 = 340Hz

c) We want to find the fifth harmonic for a string whose third overtone (this is the same as the third harmonic) frequency is 984 Hz.

Then if the frequency of the first harmonic is F, we know that:

3*F = 984 Hz

With this we can find the value of F:

F = 984 Hz/3 =328 Hz

Now that we know the frequency of the first harmonic, we can find the frequency of the fifth harmonic:

5*F = 5*328 Hz = 1640 Hz

8 0
3 years ago
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