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ivann1987 [24]
3 years ago
15

Classify the animal hydra and plant mustard with two characters of each​

Physics
1 answer:
inna [77]3 years ago
4 0

Answer:

Mustard seeds, both white and brown, are nearly globular in shape, finely pitted, odourless when whole, and pungent-tasting

Image result for character of hydra

Introduction: Hydra are inconspicuous freshwater relatives of corals, sea anemones and jellyfish. All are members of the phylum Cnidaria, characterized by radially symmetrical bodies, presence of stinging tentacles and a simple gut with only one opening (gastrovascular cavity)

Explanation:

Hydra are a genus of small, fresh-water organisms that are classified under the phylum Cnidaria.the classification of mustard is

Dicotyledons

if my answer helps you than mark me as brainliest

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With a diameter that's 11 times larger than Earth's, _______ is the largest planet.
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With a diameter that's 11 times larger than Earth's, Jupiter is the largest planet.

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Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
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Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

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