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S_A_V [24]
3 years ago
9

List some reasons why growth characteristics are more useful on agar plates than on agar slants

Physics
2 answers:
SpyIntel [72]3 years ago
5 0
Usually, in culturing of the bacteria we have a slant and then portion f it is transferred to the agar plate. The growth characteristics are more useful in the agar plates because it is where we really do the observation because bacteria in slants are still to be transferred in the agar plates. 
joja [24]3 years ago
3 0

Answer:

There are the few reasons for growth characterization are more useful on agar plates than on agar slant are discussed below.

Explanation:

Generally, agar plates are very larger and streaking out the bacteria can be possible in it. By doing this, you can find the analyzation of individual colonies of their morphology with the other uses.

  • Agar slants are usually used for the short time period of  stock cultures of bacteria.
  • Agar slants can be usually used to find the motility of bacteria by doing a single steak inoculation up the center of slant.
  • Agar plates are used used in stock cultures growing that can be refrigerated after the incubation and for several weeks it can be maintained.
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a 1000 khz am radio station broadcasts with a power of 20 kw. how many photons does the transmitter antenna emit each second
Gemiola [76]

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1 year ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

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given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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