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Hunter-Best [27]
3 years ago
7

How many Kingdoms are there? ​

Physics
2 answers:
Basile [38]3 years ago
4 0

Answer:

5

Explanation:

alex41 [277]3 years ago
3 0

Answer:

There are 5 kingdoms

Explanation:

The scheme most often used currently divides all living organisms into five kingdoms: Moneran (bacteria), Protista, Fungi, Plantae, and Animalia. This coexisted with a scheme dividing life into two main divisions: the Prokaryote (bacteria, etc.) and the Eukaryotae (animals, plants, fungi, and protists).

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Which describe the image formed by an object on the center of curvature? Check all that apply.
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The image formed by a concave  mirror with the object placed at the center of curvature is real inverted and formed at the center of curvature. Using the ray diagram a ray from the top of the object to the mirror through f then reflected parallel to the principal axis,then the ray through the center of curvature reflected through the same point both intersect at a plane through center curvature and perpendicular  to the principal axis. The point of intersection forms the top of the image and the center of curvature forms the bottom. Therefore, the correct choices are : real and inverted

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A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1
GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

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\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

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\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

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<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

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\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

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\large\textsf{\underline{Now substituting the required values}}

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\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

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The vertical position of projectile at y.

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\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

⠀

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\textsf{ \large {\underline{Now substituting the required values}}  }

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\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

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⠀

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<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>

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A solid disk of radius 5.50 cm and mass 1.25 kg , which is rolling at a speed of 1.50 m/s , begins rolling without slipping up a
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Answer:

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