A plumber has a 2-meter length of pipe. He needs to cut it into sections that are 10 centimeters long. How many sections will he
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Hey There!! ~
The answer to this is: the upper bound for the length is Lower and Upper Bounds
The lower bound is the smallest value that will round up to the approximate value.
The upper bound is the smallest value that will round up to the next approximate value.
Ex:- a mass of 70 kg, rounded to the nearest 10 kg, The upper bound is 75 kg, because 75 kg is the smallest mass that would round up to 80kg.
Here , A length is measured as 21cm correct to 2 significant figures. We need to find what is the upper bound for the length . let's find out:
As discussed above , upper bound for any number will be the smallest value in decimals which will round up to next integer value . So , for 21 :
⇒
21.5 cm on rounding off will give 22 cm . So , the upper bound for the length is
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Graph the point there.
To start this question, we should know what is the atomic number of cobalt. The atomic number (the number of protons) of Cobalt is Z =27.
Now, we know that a Cobalt 60 isotope means an isotope of Cobalt whose Atomic Mass is 60
Thus, in a Cobalt 60 isotope, the number of neutrons in the nucleus are .
From the question we know that the given nuclear mass is 59.933820 u.
Now, the mass defect of Cobalt 60 can be easily calculated by adding the masses of the protons and the neutrons as per our calculations and subtracting the given nuclear mass from it.
Thus,
Mass Defect = (Number of Protons Mass of Proton given in the question) + (Number of Neutrons
Mass of Neutron given in the question)-59.933820 u
Mass Defect =
Thus, the required Mass Defect is
In eV, the Mass Defect is