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bagirrra123 [75]
3 years ago
13

26500 in scientific notation

Physics
2 answers:
uysha [10]3 years ago
5 0

Answer:

2.56*10^4

26500

(We have to move the decimal 4 places)

2.6500*10^4

So the answer is 2.6500*10^4 or 2.56*10^4

Hope this helps! :)

ki77a [65]3 years ago
3 0

Answer:

in scientific notation 26500 is 2.65×10⁴

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Answer:

44.1613858478 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 99.4

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s

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An electron shell can hold 2(n^2) electrons (technically) where n is the shell number, i.e. shell 1 can hold 2, shell 2 can hold 8, 3 holds 18 and so on.
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In other words, you should have 2 electron shells on the atom, shell 1 with 2 e- and shell 2 with 5 e-.
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When you cook a marshmallow on a metal poker tool over an open flame, energy is transferred. Identify the three different ways t
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When we cook a marshmallow on a metal poker tool over an open flame, there are three ways in which heat energy is transferred: Conduction, convection, and radiation.

<h3>Heat energy transfer</h3>

Heat transfer is the natural transfer of heat from an object with a higher temperature to an object with a lower temperature. Heat transfer can occur in three ways, namely conduction, convection, and  radiation.

  1. Conduction occurs when heat flows from a place with a high temperature to a place with a lower temperature using a fixed heat-conducting medium. Heat transfer from the open flame to the marshmallows via direct fire contact with the marshmallows is an example of conduction.
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1 year ago
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A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running
Katena32 [7]

Answer:

1

 t_a  =  13.49 \  s

2

 The distance is   D = 55.2 \  m    

Explanation:

From the question we are told that

   The maximum speed of the cheetah is  v =  103 \  km/h =  28.61 \  m/s

    The maximum of  gazelle is  u =  76.5 \  km/h =  21.25 \  m/s

     The distance ahead is d =  99.3 \  m

Let  t_a denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

           v* t_a = d +  u* t_a

=>          28.61 t_a = 99.3 +  21.25t_a

=>          7.36 t_a = 99.3

=>         t_a  =  13.49 \  s

Now at t =  7.5 s  

             7.5 v = D+  7.5u

=>          28.61 * 7.5  = D +  21.25* 7.5

=>          7.36 *  7.5 =D

=>         D = 55.2 \  m        

Hence the for the gazelle to escape the cheetah it must be 55.2 m

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