Answer:
The correct answer is option A: they are isotopes.
Explanation:
From atom X we know that the number of protons is 7 and the number of neutrons is 7 and from atom Z we know that the number of protons is 7 and the number of neutrons is 8.
Since the number of protons of atom X and atom Z is the same, we have that atom X and atom Z is the same element. The difference in the number of neutrons tells us that atom X and atom Z are isotopes. Remember that an isotope is one element that has atoms with different numbers of neutrons.
The mass number is given by:
Where <em>n</em> is the number of neutrons and <em>p </em>is the number of protons.
For atom X and atom Z we have:

Hence, they have a different mass number.
We know that the element with 7 protons is nitrogen. The first isotope is
and the second isotope is
. Both isotopes are stables (they are not radioactive).
Therefore, the correct answer is option A: they are isotopes.
I hope it helps you!
Answer:
Heavy Rains.
These heavy rains can lead to severe flooding by oversaturating the ground, overfilling storm drains, or causing rivers to spill over their banks or levees
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 
Savanna regions developed during the Triassic period. is true
To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:
Mass of copper wire
Density
Resistively of copper 
Resistance (R) = 0.390\Omega
Volume is defined as,

(1)
We know that,


(2)
Multiplying equation we have




Therefore the length of the wire is 1.68m