Answer:
d. None of the above.
Explanation:
In a parabolic motion, you have that in the complete trajectory the component velocity is constant and the vertical component changes in time. Then, the total velocity vector is not zero.
In the complete trajectory the gravitational acceleration is always present. Then, the grasshopper's acceleration vector is not zero.
At the top of the arc the grasshopper is not at equilibrium because the gravitational force is constantly acting on the grasshopper.
Then, the correct answer is:
d. None of the above.
Answer:
181.48 N
Explanation:
Calculate the area :
Area = pi * r² ;
pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m
Area 1, A1 = 3.14 * 0.1² = 0.0314 m²
Area 2, A2 = 3.14 * 0.9² = 2.5434 m²
Force, F = mass * acceleration due to gravity
F2 = 1500 * 9.8 = 14700 N
Force 1 / Area 1 = Force 2 / Area 2
Force 1 = (Force 2 / Area 2), * Area 1
Force 1 = (14700 / 2.5434) * 0.0314
Force = 5779.6650 * 0.0314
= 181.48 N
Answer:
Explanation:
Electric field due to a point charge Q at a point at distance d is given by the relation
E = 
Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.
The charges are situated on the corners of a square in such a way that
equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.
On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero
Overall, net field due to all the four charges will be zero
Answer:
Explanation:
If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s
so here the total time of the motion above the top point of pole is given as
now we have
so this is the speed at the top of flag pole
now we have
now the height of flag pole is given as
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:
