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hram777 [196]
3 years ago
15

What are three major types of fungi and an example of each.​

Physics
2 answers:
Lelu [443]3 years ago
6 0

Answer:

Mushroom, mold, yeast

Explanation:

Nesterboy [21]3 years ago
3 0

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

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5) A 5 kg watermelon is raised 3 m by carrying it up the stairs to the second
Nutka1998 [239]

Answer: 150J

Explanation:

m = 5 kg

g = 10 m/s2

h = 3m

P.E = mgh

P.E = 5 x 10 x 3

P.E = 150J

3 0
3 years ago
A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ
ra1l [238]

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

\Delta U = q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

For a proton,

q=+e=1.6\cdot 10^{-19}C

And since \Delta V=100 V

The energy gained by the proton is

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

For an alpha particle,

q=+2e=3.2\cdot 10^{-19}C

Therefore, the energy gained is

\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

q=+e=1.6\cdot 10^{-19}C

So the energy gained is the same as the proton:

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

6 0
3 years ago
Which of these would most likely be a parts of a lab procedure?
vladimir2022 [97]
C . Record the time to complete a chemical reaction
6 0
3 years ago
Read 2 more answers
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
3 years ago
List of three questions that you are good at science questions.
Klio2033 [76]

Answer:

cycles, graphing, precise measurementation

Explanation:

7 0
3 years ago
Read 2 more answers
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