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Alisiya [41]
3 years ago
10

The organism shown above belongs to A. the plant kingdom. B. both the animal kingdom and the protist kingdom. C. the animal king

dom. D. the protist kingdom.The organism shown above belongs to A. the plant kingdom. B. both the animal kingdom and the protist kingdom. C. the animal kingdom. D. the protist kingdom.The organism shown above belongs to A. the plant kingdom. B. both the animal kingdom and the protist kingdom. C. the animal kingdom. D. the protist kingdom.

Physics
2 answers:
Bumek [7]3 years ago
6 0

The organism shown above belongs to "the animal kingdom".

<u>Answer:</u> Option C

<u>Explanation:</u>

The biological classification's hierarchy involves eight important taxonomic ranks. They are written below in decreasing order: Life > Domain > Kingdom > Phylum > Class > Order > Family > Genus > Species.

The U.S and Canada have historically used a six-kingdom scheme in their traditional textbook, such as Plantae, Animalia, Protista, Fungi, Archaea / Archaebacteria, and Bacteria / Eubacteria. The given image depicts snake which come under "Animalia Kingdom", carnivorous reptiles to  suborder Serpentes.

Korolek [52]3 years ago
6 0

Answer:

c i did it

Explanation:

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Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

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3 years ago
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