The relative atomic
mass
of boron is 10.81. Find the abundance
percentage of ¹°B and "B
is
Answers
Answer:
1200 m high. (i) how much work the sherpa
has to do against gravity? (ii) also, find the change in
potential energy of the package if the potential energy
at the base of the mountain is taken as zero.
(take g = 10 m s-2) (ans. 1.2 * 106 j, 2.4 x 1053)
at the base of the mountain is taken as zero.
(take g = 10 m s-2) (ans. 1.2 * 106 j, 2.4 x 1053)
a sherpa of mass 80 kg is instructed to carry a package
of mass 20 kg from the base camp to the summit of a
mountain 1200 m high. (i) how much work the sherpa
has to do against gravity? (ii) also, find the change in
potential energy of the package if the potential energy
at the base of the mountain is taken as zero.
(take g = 10 m s-2) (ans. 1.2 * 106 j, 2.4 x 1
of the package if the potential energy
at the base of the mountain is taken as zero.
(take g = 10 m s-2) (ans. 1.2 * 106 j, 2.4 x 1energya sherpa of mass 80 kg is instructed to carry a package
of mass 20 kg from the base camp to the summit of a
mountain 1200 m high. (i) how much work the sherpa
has to do against gravity? (ii) also, find the change in
potential energy of the package if the potential energy
at the base of the mountain is taken as zero.
(take g = 10 m s-2) (ans. 1.2 * 106 j, 2.4 x 1053)
at the base of the mountain is taken as zero.
(take g = 10 m s-2) (ans. 1.2 * 106 j, 2.4 x 1053)
a sherpa of mass 80 kg is instructed to carry a package
of mass 20 kg from the base camp to the summit of a
mountain 1200 m high. (i) how much work the sherpa
has to do against gravity? (ii) also, find the change in
potential energy of the package if the potential energy
at the base of the mountain is taken as zero.
(take g = 10 m s-2) (ans. 1.2 * 106 j, 2.4 x 1energya sherpa of mass 80 kg is instructed to carry a package
of mass 20 kg from the base camp to the summit of a
mountain 1200 m high. (i) how much work the sherpa
has to do against gravity? (ii) also, find the change in
potential
Answer:
We have,
Average atomic mass =
100
(mass× percentage abundance of 1)+(mass× percentage abundance of 2)
10.81 =
100
(11×80)+(x×20)
x=10.05≈10 (since mass number is a whole number)
Explanation:
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