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liq [111]
2 years ago
5

Which characteristic is shared by cnidarians and flatworms? A) dorsoventrally flattened bodies

Biology
1 answer:
KatRina [158]2 years ago
6 0

The correct option is (C) A digestive system with a single opening.

Which characteristic is shared by both cnidarians and Flatworms?

  • The class Soft-bodied stinging creatures like corals, sea anemones, and jellyfish are categorized as cnidaria. Numerous aquatic settings contain cnidarians. Sea anemones can be found all around the world, from the equator to the freezing northern regions and from small tide pools to the deep ocean floor.
  • Simple flatworm-like creatures belong to the phylum Platyhelminthes. Flatworms are parasites that live in or on other animals and are found on land, in fresh water, and in the ocean (e.g., tapeworms). Flatworm parasites can harm or even kill the host organism when they dwell on or inside of other species, including humans.
  • Although flatworms, like cnidarians, have a digestive system with only one opening into the digestive cavity, the cavity in independently living marine flatworms branches into every section of the body.

Learn more about the Phylum platyhelminthes with the help of the given link:

brainly.com/question/14441797

#SPJ4

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What is the magnitude of the force exerted by the biceps fbiceps? What is the magnitude of the force exerted by the elbow felbow
Slav-nsk [51]

Answer:

<u> </u><u>The magnitude of the force exerted by the biceps </u>f_b_i_c_e_p_s<u> is 218.38N</u>

<u>The magnitude of the force exerted by the elbow </u>f_e_l_b_o_w<u> is 194.37N</u>

Explanation:

Firstly , lets calculate the magnitude of force exerted by biceps f_b_i_c_e_p_s -

Mass of the forearm=

m_f=1.50kg

Mass of the ball =

m_b_a_l_l=950g

   = 0.950kg

so, weight of the forearm=

w_f=m_f g

weight of the ball=

w_b_a_l_l=m_b_a_l_l g

Now , balancing torque about the elbow , we have-

F_b_i_c_e_p_s\times d_b_i_c_e_p_s= w_f\times\frac{d_b_a_l_l}{2} + w_b_a_l_l\times d_b_a_l_l

(here , distance of the bicep is 2.50cm = 0.025m and distance of the ball is 36cm= 0.36m)

Now putting the given values in the formula -

F_b_i_c_e_p_s\times 0.025= 0.95\times9.8\times\frac{0.36}{2} +1.50\times 0.36

F_b_i_c_e_p_s=218.38N

Now, lets calculate the force exerted by elbow F_e_l_b_o_w -

Balancing the vertical forces

F_b_i_c_e_p_s=F_e_l_b_o_w+w_f+w_b_a_l_l

F_e_l_b_o_w = 218.38-(1.00\times9.8)-(0.95\times9.8)

F_e_l_b_o_w=194.37N

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What products from the ""light"" reactions are used in the ""dark"" reactions?
liubo4ka [24]

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The products are organic sugar molecules

Explanation:

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