The wood thrush is a bird species that lives in deciduous forests in north america. wood thrushes nest in tall trees, forage for insects and snails in decaying leaf litter on. the forest floor, and drink water from nearby fresh water sources. indentify one biotic factor and one abiotic factor critical to this speicies survival

In ecology and biology, the living components of an ecosystem are known as biotic factors. They are distinguished into three groups, that is, consumers of heterotrophs, producers or autotrophs, and decomposers or detritivores. On the other hand, the abiotic factors or components refers to the non-living physical or chemical parts of the environment, which influences living species, and the working of the ecosystems.  

In the given case of wood thrush, the biotic factor would be the supply of food in the form of snails and insects, and the abiotic factor would be the supply of water. Both these factors would be critical for the survival of the species.  

out of these options:

ability to burn paper

pure sodium reacting violently with water

the boiling point of water

helium does not tend to react with anything

i would say this is a physical property:

helium does not tend to react with anything

why? :

i said helium does not tend to react with anything because well, it reacts to ! but ability to burn paper, pure sodium reacting violently with water, & the boiling point of water are chemical or

physical properties: the properties of a substance that can be observed without changing the chemical makeup of the substance

chemical properties: properties of a substance relating to the chemical nature and reactivity of the substance

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j