It would be d all of the above
Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.
Magnetic field are a region around a magnetic material or a moving electric charge within which the force of magnetism acts. Magnetic fields are produced by moving electric charges. Everything is made up of atoms, and each atom has a nucleus made of neutrons and protons with electrons that orbit around the nucleus. Since the orbiting electrons are tiny moving charges, a small magnetic field is created around each atom.
Similarities between magnetic fields and electric fields: Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like pole repel unlike poles attract. Electric field points in the direction of the force experienced by a positive charge.
Answer:

Yes it is better to pull the rope rather than push it
Explanation:
Let the force is applied at an angle of 60 degree
so we will have net vertical force on the crate is given as

here we know




now friction force on the crate is given as




Yes it is better to pull the rope rather than push it
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:
Answer:
19.5°
Explanation:
The energy of the mass must be conserved. The energy is given by:
1) 
where m is the mass, v is the velocity and h is the hight of the mass.
Let the height at the lowest point of the be h=0, the energy of the mass will be:
2) 
The energy when the mass comes to a stop will be:
3) 
Setting equations 2 and 3 equal and solving for height h will give:
4) 
The angle ∅ of the string with the vertical with the mass at the highest point will be given by:
5) 
where l is the lenght of the string.
Combining equations 4 and 5 and solving for ∅:
6) 