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:
![I = \frac{V}{R} (1 - e^{- \frac{R}{L}t } )](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D%20%281%20-%20e%5E%7B-%20%5Cfrac%7BR%7D%7BL%7Dt%20%7D%20%29%20)
For the current to reach half its final value:
Yes the lightest one will go the highest. if 2 objects with the same mass like a kilogram of feathers and a kilogram of steel they both would hit the ground at the same time bec they are both a kilogram
Answer:
ω = 2.1 rad/sec
Explanation:
- As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
- This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
- Now, we need to ask ourselves: what supplies this force?
- In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
- We know that this force can be expressed as follows:
![f_{frs} = \mu_{s} * F_{n} (1)](https://tex.z-dn.net/?f=f_%7Bfrs%7D%20%3D%20%5Cmu_%7Bs%7D%20%2A%20F_%7Bn%7D%20%281%29)
where μs = coefficient of static friction between the rock and the merry-
go-round surface = 0.7, and Fn = normal force.
- In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
- Fn = m*g (2)
- This static friction force is just the same as the centripetal force.
- The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:
![F_{c} = m* \omega^{2}*r (3)](https://tex.z-dn.net/?f=F_%7Bc%7D%20%3D%20m%2A%20%5Comega%5E%7B2%7D%2Ar%20%283%29)
- Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:
![\omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Cmu_%7Bs%7D%20%2A%20g%7D%7Br%7D%20%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.7%2A9.8m%2Fs2%7D%7B1.6m%7D%7D%20%3D%202.1%20rad%2Fsec)
- This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.
A) The stone moves along the vertical direction by unifom accelerated motion, with acceleration equal to g (gravitational acceleration), starting from initial position h above the ground and with initial velocity equal to zero. So, its vertical position follows the law:
![y(t) = h - \frac{1}{2}gt^2 = 450 - \frac{1}{2}(9.8)t^2 [m] = 450 - 4.9 t^2 [m]](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20-%20%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20%3D%20450%20-%20%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2%20%5Bm%5D%20%3D%20450%20-%204.9%20t%5E2%20%5Bm%5D%20%20)
b) The time the stone takes to reach the ground is the time t at which its vertical position y(t) becomes zero:
![0=y(t) =450-4.9 t^2](https://tex.z-dn.net/?f=0%3Dy%28t%29%20%3D450-4.9%20t%5E2)
and if we solve it, we find
![t= \sqrt{ \frac{450}{4.9} }=9.6 s](https://tex.z-dn.net/?f=t%3D%20%5Csqrt%7B%20%5Cfrac%7B450%7D%7B4.9%7D%20%7D%3D9.6%20s%20)
c) Since it is a uniform accelerated motion, the velocity of the stone at time t is given by
![v(t) = v_0 -gt=-gt](https://tex.z-dn.net/?f=v%28t%29%20%3D%20v_0%20-gt%3D-gt)
where the initial velocity is zero:
![v_0 = 0](https://tex.z-dn.net/?f=v_0%20%3D%200)
. The stone hits the ground at t=9.6 s, so its velocity at that time is
![v(9.6s)=-gt=-(9.81 m/s)(9.6 s)=-94.2 m/s](https://tex.z-dn.net/?f=v%289.6s%29%3D-gt%3D-%289.81%20m%2Fs%29%289.6%20s%29%3D-94.2%20m%2Fs)
where the negative sign means it is directed downward.
d) In this case, since the initial velocity is not zero, the position at time t is given by
![y(t) = h -v_0 t - \frac{1}{2}gt^2 = 450 - 5t -4.9t^2 [m]](https://tex.z-dn.net/?f=y%28t%29%20%3D%20h%20-v_0%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20%3D%20450%20-%205t%20-4.9t%5E2%20%5Bm%5D%20)
where
![v_0=5 m/s](https://tex.z-dn.net/?f=v_0%3D5%20m%2Fs)
is the initial velocity.
The time the stone takes to reach the ground is the time t such that y(t)=0, so we have:
![450-5t-4.9t^2 =0](https://tex.z-dn.net/?f=450-5t-4.9t%5E2%20%3D0)
and by solving this equation, we find