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eimsori [14]
3 years ago
13

WILL GIVE BRAINLIEST AND 30 POINTS!

Physics
1 answer:
mariarad [96]3 years ago
8 0

Answer:

Gallium

Explanation:

Gallium is one such element used as a do/pant in a p-type semiconductor.

A do/pant is an impurity added to a semi-conductor used to alter its properties. Semi-conductors have a wide range of applications. They will conduct heat and electricity only under certain conditions. This property is highly desirable and find a wide application in electronics.

For p-type conductors, they are best do/ped with elements with 3 valence electrons. These are group 3 elements. From the choices, only gallium belongs to this group.

Other elements given are good do/pants for n-type semiconductors. They have 5 valence electrons.

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How does radioactive decay work
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Two scientists interpret the same set of data differently. How would the scientific community deal with this problem?
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One cereal bar has a mass of 37 g. What is the mass of 6 cereal bars? Is that more than or less than 1 kg? Explain your answer.
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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
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