Answer:
E) True. The girl has a larger tangential acceleration than the boy.
Explanation:
In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.
Angular and linear quantities are related
v = w r
a = α r
the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m
as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different
v₁ = w 1,2 (boy)
v₂ = w 1.8 (girl)
whereby
v₂> v₁
reviewing the claims we have
a₁ = α 1,2
a₂ = α 1.8
a₂> a₁
A) False. Tangential velocity is different from zero
B) False angular acceleration is the same for both
C) False. It is the opposite, according to the previous analysis
D) False. Angular acceleration is equal
E) True. You agree with the analysis above,
Answer:
Honestly i think the answer is B
Explanation:
Answer:
D
Explanation:
appearance is not a imp factor . location could be imp. becoz a proper environment is need to study such courses.
Well formation of metallic bond depends on free electrons.smaal sized atoms hold their electrons more firmly as compared to large size atoms ,this z due to distance of outer shell electrons by nucleus..in this way no of free electrons affect strength of metallic bond..smaal sized atoms release less free electrons..
Answer:
In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.
Explanation:
The meaning of odd nuclei is atomic mass is odd.
A=odd number.
A=Z+n
Here, Z is proton either it will odd or n will odd which is neutron.
Now according to the shell model the left out proton or neutron will contribute to the spin and parity.
For example,
Take the case of isotope of nitrogen-15.
Here Z is 7, and n is 8 will not contribute in spin.
Now, for Z=7.

Here,

and, L=1.
Fort parity,

Put the value of L.
Parity will be -1.
Now, spin will be
.