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Papessa [141]
3 years ago
11

Rooted plants are most likely found in which aquatic zone?

Physics
1 answer:
Mnenie [13.5K]3 years ago
3 0

Ans. Littoral zone.

Littoral zone can be defined as the area of a rive, sea, or lake, which lies close to the shore. It is characterized by abundant sunlight, dissolved oxygen, nutrients, and high water motion. These characteristics make it suitable for   marine life. Hence, littoral zone is enriched with various plants and animal species, and coral reefs.  

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Can toe touches help you improve flexibility? Yes/No
jolli1 [7]
Yes it helps improve flexibility
7 0
3 years ago
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When comparing saturated and naturally occurring unsaturated fats, the unsaturated fats have __________ and are __________ at ro
Len [333]

Answer: have "cis C=C double bonds" and "liquid" at room temperature.

Explanation:

The unsaturated fatty acids have one or more C=C double bonds in the cis formation. Thus, this results in the molecules not been as stable as the saturated fats. They have weaker intermolecular bonds thus resulting in lower melting point . The consequently results in it being liquid at room temperature.

5 0
4 years ago
given that gravitational is represented by the formula f=gm1,m2/r separated by R determine the basic units of constant G​
Cloud [144]

Answer:

G is the Universal Gravitational Constant (equal to: 6.672 x 10-11Nm2/kg2).

6 0
2 years ago
A 17,250 kg rocket is pushed with a thrust of 6,450,284 N. What is the acceleration of the rocket?
Leni [432]

Data given:

m=17,250 kg

F=6,450,284 N

a=?

Formula needed:

F=ma

Solution:

a= F/m

a=373,9295 m/s²

8 0
3 years ago
A child is sitting on the seat of a swing with ropes 5 m long. Her father pulls the swing back until the ropes make a 30o angle
morpeh [17]

Answer:

v = 3.7 m/s

Explanation:

As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:

m. g. h = 1/2 m v²

The only unknown (let alone the speed) in the equation , is the height from which the swing is released.

At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.

As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:

h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m

Replacing in the energy conservation equation, and solving for v, we get:

v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s

7 0
4 years ago
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