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White raven [17]
3 years ago
10

A forest is an ecosystem. There are many plants and animals in the forest. What is true about the plants and animals? The plants

depend on the animals. The animals depend on the plants. The plants do not depend on the animals. The animals do not depend on the plants.
Physics
2 answers:
Volgvan3 years ago
7 0

Answer:

The plants depend on animals and animals depend on plants.

Plants depend on animals for nutrients, pollination and seed dispersal. Animals are totally dependent on plants. First of all plants produce oxygen which is vital for life on earth. Then, for many animals plants are their basic food, or if plants don’t make up 100 % of their diet, plants are nevertheless essential in their food supply.

ivanzaharov [21]3 years ago
6 0

Answer:

A. The plants depend on the animals.

B. The animals depend on the plants.

Explanation:

Plants require animals to spread their seeds around the area, also known as seed dispersal. They also use the help of other animals for pollination. Bees can be used as a great example for pollination and so as butterflies, flies, etc. Some plants have seeds with barbs that stick to the animals fur. Some just simply stick to the animals paws and/or fur until dropping off to another place. Other seeds are consumed and then composed. Here's an example. A deer eats an apple. Everyone knows that apples have seeds, hence, the deer consumes the seeds. The deer will then compose the digested food along with the seeds. Now, the apple's seeds have settled in another place and will start its growth. Animals also depend on the plants. Plants are a source of food for many herbivores and omnivores. Plants also have a high nutrition, giving animals the nourishment needed. They also produce oxygen and take in carbon dioxide. This is necessary for earth and its wildlife on it.

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yanalaym [24]

The information that could be gathered about a star whose light curve has multiple symmetrical depths is ; The shape and surface variegation of the star

The light curves of a KBO ( moons and stars ) are measured as a rate of the brightness of a star in relation to time. therefore the study of the light curve having multiple symmetrical depths ( depth of brightness ) will give an information about the shape/size and the surface variegation of the star

Hence we can conclude that The information that could be gathered about a star whose light curve has multiple symmetrical depths is ; The shape and surface variegation of the star

Learn more : brainly.com/question/19573734

7 0
2 years ago
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A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before
Darina [25.2K]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before  \: it \: comes \: to \: rest \:( s_{2} )}

\\

{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity   = v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \:  s_{1} \: penetration =  \dfrac{v}{2}  \:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{s_{1} =  \dfrac{3}{100}  = 0.03 \: m}

\\

☯ As we know that,

\\

\dashrightarrow\:\: \sf{ {v}^{2}  =  {u}^{2} + 2as }

\\

\dashrightarrow\:\: \sf{  \bigg(\dfrac{v}{2} \bigg)^{2}  =  {v}^{2}   + 2a s_{1}}

\\

\dashrightarrow\:\: \sf{  \dfrac{ {v}^{2} }{4}  =  {v}^{2}  + 2 \times a \times 0.03  }

\\

\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4}  -  {v}^{2}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{\dfrac{ -  3{v}^{2} }{4}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{a =  \dfrac{ - 3 {v}^{2} }{4 \times 0.06}  }

\\

\dashrightarrow\:\: \sf{ a =  \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }

\\

\:\:\:\:\bullet\:\:\:\sf{  Initial\:velocity=v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }

\\

\dashrightarrow\:\: \sf{  {v}^{2}  =  {u}^{2}  + 2as}

\\

\dashrightarrow\:\: \sf{{0}^{2}  =  {v}^{2}  + 2 \times  \dfrac{ - 25 {v}^{2} }{2}  \times s  }

\\

\dashrightarrow\:\: \sf{ -  {v}^{2}  =  - 25 {v}^{2}  \times s  }

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{ -  {v}^{2} }{ - 25 {v}^{2} }}

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{1}{25} }

\\

\dashrightarrow\:\: \sf{ s = 0.04 \: m }

\\

☯ For left penetration (s₂)

\\

\dashrightarrow\:\: \sf{s =  s_{1} +  s_{2}  }

\\

\dashrightarrow\:\: \sf{  0.04 = 0.03 +  s_{2}}

\\

\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }

\\

\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}

\\

\star\:\sf{Left \: penetration \: before  \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\

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2 years ago
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insens350 [35]

Answer:

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Explanation:

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A 15 kg box sits still on a rough inclined plane (no math required here).
nexus9112 [7]

Answer:

Explanation:

The force of friction equals the sine component of the force due to gravity

3 0
2 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

8 0
3 years ago
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