1) yeasts example is Sacchromyces Cerevisiae which is a baker's or brewer's yeast
2) molds example is Rhizopus a type of mold that appears on old bread
3) mushrooms example is Amanita Phalloides also known as the "Death Cap " is a very poisonous mushroom and should not be ingested
The momentum increases by a factor of 2
Explanation:
We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.
The kinetic energy of the rocket is:
(1)
where
m is the mass
v is the velocity
The momentum of the rocket is
(2)
From eq.(1) we get

and substituting into (2),

Now in this problem we have:
- The kinetic energy of the rocket is increased by a factor 8:

- The mass is reduced by half:

Substituting, we find the new momentum:

So, the momentum increases by a factor of 2.
Learn more about momentum and kinetic energy:
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Answer:
A. Remove everything in the refrigerator to lighten the load.
B. Put a lubricant between the surface of the object and the floor
C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily
Explanation:
Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.
F = μN where N is the normal force which depends on the mass.
Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.
Answer:
Hence the pressure is 
Explanation:
Given data
Q=1500 J system gains heat
ΔV=- 0.010 m^3 there is a decrease in volume
ΔU= 4500 J internal energy decrease
We know work done is
W= Q- ΔU
=1500-4500= -3000 J
The change in the volume at constant pressure is
ΔV= W/P
there fore P = W/ΔV= -3000/-0.01= 3×10^5
Hence the pressure is 