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Tems11 [23]
3 years ago
5

Name three types of variables in an experiment.

Physics
2 answers:
Mandarinka [93]3 years ago
8 0
These changing quantities are called variables. A variable is any factor, trait, or condition that can exist in differing amounts or types. An experiment usually has three kinds of variables: independent, dependent, and controlled.
makvit [3.9K]3 years ago
7 0

Answer:

There are three main variables: independent variable, dependent variable and controlled variables. Example: a car going down different surfaces

Explanation: pls mark my answer as brainlist

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If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial accelerat
Cerrena [4.2K]

Answer:

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

F=\frac{1}{4(\pi)Eo }\frac{|qp*qe|}{r^{2} }\\   F=(8.988*10^{9}Nm^{2}/C^{2})*\frac{(1.602*10^{-19})^{2}  }{(6.50*10^{-10} )^{2}  } \\F=5.46*10^{-10}N

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

a=F/Me\\a=(5.46*10^{-10} )/9.109*10^{-31}\\ a=5.993*10^{20}m/s^{2}

For(b) Acceleration of proton

a=F/Mp\\a=(5.46*10^{-10} )/1.673*10^{-27} \\a=3.264*10^{17}m/s^{2}

3 0
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