Answer:
Nucleus A with 7 protons and 7 neutrons and Nucleus C with 7 protons and 5 neutrons are isotopes of the same elements
Explanation:
Isotopes are elements that have the same atomic structure but different molecular structure. An atom that has the same atomic number but different mass number are known to be isotopes.
The proton of an atom is the same as its atomic number while the sum of the proton and neutron is equal to its mass number.
According to the question, nuclei that has the same number of proton are isotopes of the same element. Therefore nuclei A and C with 7 protons each are isotopes of the same element since they have the same atomic number i.e number of proton = atomic number.
Their atomic masses of nuclei A and C are 14 and 12 respectively
The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.
Equilibrium to the crest . . . that's the amplitude.
Crest to trough . . . that's double the amplitude.
Trough to trough . . . How did that get in here ? Yes, that's
the wavelength, but it has nothing to do
with vertical displacement.
Frequency . . . that's how many complete waves pass a mark
on the ground every second. Doesn't belong here.
Notice that this has to be a transverse wave. If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.
It also has to be a vertically 'polarized' wave. If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement. It still has an amplitude, but the amplitude
is all horizontal.
Answer:
The frequency of the oscillations in terms of fo will be f2=fo/3
E xplanation:
T= 
=1:3
⇒f2=fo\3
Here frequency f is inversely poportional to square root of mass m.
so the value of remainder of frequency f2 and fo is equal to 1:3.
⇒
= 
⇒
= 1:3
⇒f2=
I would say that it would take her 35 * 2 cashing Bill properly because I multiply 0.25 times 16 which gave me 1.50 + 2.50 equals 3.50
1. Velocity at which the packet reaches the ground: 121.2 m/s
The motion of the packet is a uniformly accelerated motion, with constant acceleration 
directed downward, initial vertical position 
, and initial vertical velocity 
. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:
substituting, we find
2. height at which the packet has half this velocity: 562.6 m
We need to find the heigth at which the packet has a velocity of
In order to do that, we use again the same SUVAT equation substituting 
with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:
Which means that the heigth of the packet was
