Answer:
1) Where
2) How far
3) How far, in what direction
4) How fast
5) How fast,in what direction
Answer:
Once a carnivorous plant has procured an item for dinner, it has to have some way to turn it into fertilizer. What carnivorous plants do is very similar to what humans do with their dinner after they have eaten it. Most carnivorous plants have glands that secrete acids and enzymes to dissolve proteins and other compounds. The plants may also enlist other organisms to help with digestion. The plants then absorb the nutrients made available from the prey.
Drosera releases digestive juices through the glands at the tip of its tentacles and absorbs the nutrients through the tentacles, leaf surface, and sessile glands. In order to do this it bends its tentacles and rolls or bends the leaf to get as many tentacles as possible into contact with the prey for digestion and to make as much leaf surface available for absorption. Its relative Drosophyllum has differently structured, non moving tentacles and doesn't use them directly for digestion. Instead it has specialized glands on the surface of the leaf that release the digestive enzymes (see Carniv. Pl. Newslett. 11(3):66-73 ( PDF ) for drawings and discussion).
The sealed trap of Dionaea does digestion in a way similar to the leaf surface digestion carnivores—upon capture of a prey, digestive enzymes in mucous are released. The advantage of the sealed trap of Dionaea is rain won't wash away the nutrients as digestion proceeds.
The sealed trap carnivores Aldrovanda and Utricularia already have water in their traps so they only need to release enzymes. Utricularia appears to release the enzymes continuously into its traps.
The other carnivorous plants use either a mixed mode of digestive enzymes and partner organisms (Genlisea, Sarracenia, most Nepenthes, Cephalotus, some Heliamphora, Roridula) or other organisms exclusively for digestion (most Heliamphora, some Nepenthes, Darlingtonia). Part of the reason for partnering with other organisms is that the plants actually have little choice in the matter. This could also be a factor for the leaf surface and sealed trap digesters as well. The prey will have gut flora that are quite capable of digesting their host when it dies. In addition, insect larvae, frog tadpoles, and predacious protozoans will or will attempt to take up residence in water-filled traps. The plant releasing digestive enzymes and acids into the traps will help tip the nutrition balance to themselves, but there are limits.
Explanation:
Explanation:
Given Data
Total mass=93.5 kg
Rock mass=0.310 kg
Initially wagon speed=0.540 m/s
rock speed=16.5 m/s
To Find
The speed of the wagon
Solution
As the wagon rolls, momentum is given as
P=mv
where
m is mass
v is speed
put the values
P=93.5kg × 0.540 m/s
P =50.49 kg×m/s
Now we have to find the momentum of rock
momentum of rock = mv
momentum of rock = (0.310kg)×(16.5 m/s)
momentum of rock =5.115 kg×m/s
From the conservation of momentum we can find the wagons momentum So
wagon momentum=50.49 -5.115 = 45.375 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
Speed of wagon=45.375/(93.5-0.310)
Speed of wagon= 0.487 m/s
Throwing rock backward,
momentum of wagon = 50.49+5.115 = 55.605 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
speed of wagon = 55.605 kg×m/s/(93.5kg-0.310kg)
speed of wagon= 0.5967 m/s
Answer: 5.97, 24
m = 5.97
n = 24
SN = 5.97 × 10^24 kg
Explanation:
Scientific notation or standard form is a way of rewriting very large or very small (minute) numbers in conveniently in decimal form. The standard way of writing scientific notation is of the form;
m × 10^n
To derive m, we need to reduce it to a number between 1 and 10 ( 10 exclusive)
For the case above ;
m = 5.97
To get the value of n, we need to count the number of digits from the right egde of the data to the point where we put the decimal.
For the case above;
n = 24
Therefore, the scientific notation of the mass of the earth is ;
M = 5.97 × 10^24 kg