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Vadim26 [7]
3 years ago
12

Which characteristics belong to Whittaker’s classification system? Select two options.

Physics
2 answers:
lubasha [3.4K]3 years ago
5 0
B and E sorry if I’m wrong have a good day
Lana71 [14]3 years ago
4 0

Answer:

Five kingdoms & Seven levels

Explanation:

Robert H. Whittaker proposed five kingdom classical system for living organisms in 1969. These were: Kingdom Monera, Kingdom Protista, Kingdom Fungi, Kingdom Animalia, and Kingdom Plantae. His work was the improvement of earlier work conducted by Carl Linnaeus who proposed three kingdom classification system (Animalia, Plantae, and Protista).

Whittaker's system further categorized each kingdom to seven levels. These were: Kingdom → Phylum  →  Class  →   Order  →   Family →   Genus  →   Species. This categorization was based on similarities in organisms at different levels.

PS: The scientific name of each organism is the combination of genus and family name.

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A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla
kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v

Put the value into the formula

P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}

P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}

P=0.350\ Watt

Hence, The power output of the oscillator is 0.350 watt.

3 0
3 years ago
The position x of a particle moving along x axis varies with time t as x = Asin(wt) , where A and w are constants . The accelera
Elodia [21]

Let's see

\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)

Now we know the formula of acceleration

\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)

\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2

  • From eq(1)

\\ \rm\Rrightarrow \alpha=-x\omega^2

Or

\\ \rm\Rrightarrow \alpha=-\omega^2x

6 0
2 years ago
Read 2 more answers
A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. When the temperature of th
Nesterboy [21]

Answer:

Option C is the correct answer.

Explanation:

By Charles's law we have

        V ∝ T

That is

       \frac{V_1}{T_1}=\frac{V_2}{T_2}

Here given that

      V₁ = 0.20 cubic meter

      T₁ = 333 K

      T₂ = 533 K

Substituting

      \frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\\frac{0.20}{333}=\frac{V_2}{533}\\\\V_2=\frac{0.20}{333}\times 533=0.3198m^3

New volume of the gas  = 0.3198 m³

Option C is the correct answer.

7 0
4 years ago
What is the gravitational potential energy of a 1.02 kg book on the top of a 2.0m locker
zmey [24]

Answer:

20.4

Explanation:

6 0
3 years ago
A 600 N force acts on an object with a mass of 50 kg. What is the resulting acceleration of the object?
DochEvi [55]

Answer:

<h3>The answer is 12 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{600}{50}  =  \frac{60}{5}  \\

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

4 0
3 years ago
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