HELP! IDENTIFY AND DRAW A CONCLUSION FOR THIS PROBLEM PLS! ASAP.

1.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the mass defect of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev 2.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the binding energy of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev 3.)the isotope cobalt-60 has a nuclear mass of 59.933820 u calculate the binding energy per nucleon of cobalt-60 using the following information. mass of proton: 1.007825 u mass of neutron: 1.008665 u 1 u = 931.5 mev

Assume you measured the mass of the cart to be (500 ± 1) g and the mass of the additional mass you put on the cart to be (500 ± 1) g as well. since the scale you are using in the lab cannot measure objects heavier than 600g you will have to sum up individual pieces and propagate the error. so what would be the mass and the standard error of the cart and the mass

Which describes increasing the efficiency of energy resource