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3 years ago

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What is the function of the phytoplankton in ocean ecosystems?

1 answer:

vesna_86 [32]3 years ago

5 0

Ok ok! It's A. Producer

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1. Which wave phenomenon is illustrated by this image?

Arisa [49]

Answer:

Explanation: O B. Diffraction

7 0

2 years ago

Read 2 more answers

Convert 285,000 milliliters to decaliters.

Rudiy27

The answer is 28.5 decaliters. Hope this helps

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3 years ago

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Think of a time where you have observed someone's behavior and been encouraged to do the same. What did you do? Was it a good or

ololo11 [35]

Answer:

i'm not sure if you are asking as a personal question or a book question so i'm taking it personal.

Explanation:

I was doing a simple task that was handed to me to test my responsibility and I agreed (knowing i am responsible :3). my first thought was "man , this is easy!" but then i started seeing the other kids slaking off and quiting their tasks. I thought that was against the rules, but then i saw my bff doing it too and i thought "this should be ok then!" so i did the same. other kids where still doing it. the teacher came, saw the ones still working and smiled... but when the teacher looked at the ones slaking off omg... his face was like * im gonna kill yall* we took one big gulp and whined. the teacher awarded the ones who completed the task... the others , we had to do our original task but doubled... for 3 weeks!!! it was awful!!!

I WOULD NEVER DO THAT AGAIN!!!

5 0

2 years ago

A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.

vodomira [7]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as

$f_d= f_s \frac{(v+v_d)}{(v-v_s)}$

Here,

$f_d$ =frequency received by detector

$f_s$ =frequency of wave emitted by source

$v_d$ =velocity of detector

$v_s$ =velocity of source

v=velocity of sound wave

Replacing we have that,

$f_d = 400(\frac{(343+18)}{(343-35)})$

$f_d=422 Hz$

Therefore the frequencty that will hear the passengers is 422Hz

8 0

3 years ago

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

krok68 [10]

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

3 0

3 years ago