<span>The three major types of
symbiosis are mutualism, where both species benefit, commensalism, where
one species benefits and the other is unaffected, and parasitism, where
one species benefits and the other is harmed. Symbiotic relationships can occur within an organism's body or outside of it. </span><span>Examples of mutualism include the
relationship between single-celled organisms or animals that incorporate
algae into their bodies. They give the algae necessary nutrients, and
in return receive chemical energy from the photosynthetic algae. Animals
that have this sort of relationship include some sponges, sea anemones
and clams.
Examples of commensalism include remora fish attaching to the bodies
of sharks and eating scraps of food that escape their jaws, and
barnacles living on the jaws of whales with a similar feeding strategy.
Plants have commensal relationships as well, such as many orchids that
grow on taller plants and benefit from the additional sunlight they
obtain, without actually stealing nutrients from the host plant.
Parasitic relationships are many, and parasites include all
disease-causing organisms. This category also includes insects such as
fleas that suck the blood of hosts externally. Parasitism is a very
efficient strategy for organisms, and parasites often lose many of the
features of non-parasitic life forms, instead relying on their hosts for
many of the functions of life.</span>
Answer:
a). Single replacement.
Explanation:
Because one element replaces another element in a compound
Answer:
0.231 m/s
Explanation:
m = mass attached to the spring = 0.405 kg
k = spring constant of spring = 26.3 N/m
x₀ = initial position = 3.31 cm = 0.0331 m
x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m
v₀ = initial speed = 0 m/s
v = final speed = ?
Using conservation of energy
Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy
(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²
m v₀² + k x₀² = m v² + k x²
(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²
v = 0.231 m/s
The solid, liquid and gas phases of water would have the same structure of the molecules since they are same substance. The only difference would be the distances of the molecules in the container. For a ice, the molecules are close to each other where the molecules vibrate only in place. For liquid, the molecules are freely moving and are at some distance with each other but not that far away with each other. Steam, on the other hand, would have molecules that are very far from each other and are freely moving in the whole container. As the container is heated, the size of the molecules would not change. It is only the volume that has changed. Also, the mass is the same since there is no outflow of the substances.
Answer:
The tension is 
Explanation:
The free body diagram of the question is shown on the first uploaded image From the question we are told that
The distance between the two poles is 
The mass tied between the two cloth line is 
The distance it sags is 
The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline
Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium
And this can be mathematically represented as

To obtain
we apply SOHCAHTOH Rule
So 
![\theta = tan^{-1} [\frac{opp}{adj} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7Bopp%7D%7Badj%7D%20%5D)
![= tan^{-1} [\frac{1}{7}]](https://tex.z-dn.net/?f=%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B7%7D%5D)





