Answer:
E- The star becomes a red giant (LATEST STAGE)
F- The surface of the star becomes brighter and cooler
C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand
A- The shell of hydrogen surrounding the star's nonburning helium core ignites.
D- The star's non burning helium core starts to contract and heat up
B- Pressure in the star's core decreases (EARLIEST STAGE)
(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.
Answer:
A. respiration.
Explanation:
Cellular respiration can be defined as a series of metabolic reactions that typically occur in cells so as to produce energy in the form of adenosine triphosphate (ATP). During cellular respiration, high energy intermediates are created that can then be oxidized to make adenosine triphosphate (ATP). Therefore, the intermediary products are produced at the glycolysis and citric acid cycle stage.
Additionally, mitochondria provides all the energy required in the cell by transforming energy forms through series of chemical reactions; breaking down of glucose into Adenosine Triphosphate (ATP) used for providing energy for cellular activities in the body of living organisms.
Basically, oxygen goes into the body of a living organism such as plants, humans and animals when they breathe while glucose is absorbed by the body when they eat.
Hence, the conversion of sugar to energy in the presence of oxygen is respiration.
Answer:
h = height of the hotel room from the ground floor = 237.4m
Explanation:
Change in Potential Energy of tourist = ΔPE = PE2 – PE1 = mgh
PE1 is the potential energy of tourist at the ground floor
PE1 is the potential energy of tourist at the top (hotel room)
Given
PE1 = − 2.01 × 10⁵ J
PE2 = 0J
PE2 – PE1 = mgh
0 – (− 2.01 × 10⁵ J) = mgh
2.01 × 10⁵ J = 86.4×9.8×h
h = 2.01 × 10⁵/(86.4×9.8) = 237.4m
In this problem we have the electric field intensity E:
E = 6.5 × 
newtons/coulomb
We have the magnitude of the load:
q = 6.4 × 
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × 
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × 
)(6.5 × )(1.2 × )
PE = 5.0 x 
joules
None of the options shown is correct.
Answer:
The speed of the block is 4.96 m/s.
Explanation:
Given that.
Mass of block = 1.00 kg
Spring constant = 500 N/m
Position 
Coefficient of friction = 0.350
(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less
Using formula of kinetic energy and potential energy
Put the value into the formula
Hence, The speed of the block is 4.96 m/s.
