Question

What units of measurement do scientists use $?$

Answer 1

Hint : All the scientist follows the system of units$(SI)$. Based on system of units there are seven base units metre$(m)$, kilogram$(Kg)$, second$(s)$, ampere$(A)$, candela$(cd)$, mole$(mol)$, kelvin$(K)$.

Complete Step By Step Answer:
All the scientist use the international system of units$(SI)$:
The SI is a system based upon seven base units:
 metre$(m):$length$ = $The meter is as of now characterized as the path travelled by light in a vacuum in$1299792458$of a second.
kilogram$(kg):$ mass$ = $The kilogram is the base unit of mass in the International System of Units, the current metric system, having the unit symbol $Kg$.
second$(s):$time$ = $the second was known as a "second minute", meaning the second minute division of an hour. The first division was known as a "prime minute" and is equivalent to the minute we know today. Third and fourth minutes were sometimes used in calculations.
ampere$(A):$electric current$ = $ the ampere is a measure of the amount of electric charge passing a point in an electric circuit per unit time, or one coulomb per second constituting one ampere
candela$(cd):$luminous intensity$ = $Candela, unit of luminous intensity in the International System of Units, defined as the luminous intensity in a given direction of a source that emits monochromatic radiation of frequency $540 \times {10^{12}}hertz$ .
mole $(mol):$amount of substance$ = $The mole is the unit of measurement for amount of substance in the International System of Units. It is defined as exactly $6.02214076 \times {10^{23}}$ particles, which may be atoms, molecules, ions, or electrons.
kelvin $(K):$temperature$ = $The kelvin is a base SI unit of measurement, since 2018 defined by setting the fixed numerical value of the Boltzmann constant k to $1.380649 \times {10^{ - 23}}j{k^{ - 1}}$.

Note :
Derived units are formed by various combinations of the base units.For example, velocity is defined as distance per unit of time, which in $SI$ has the dimensions of metres per second $(\dfrac{m}{s})$.