Answer:

Explanation:
We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

where
F is the net force on the object
m is its mass
a is its acceleration
In this problem:
F = 40 N is the force on the object
m = 2 kg is its mass
Therefore, the acceleration of the object is

Answer:
A.2.95 m
B.7
Explanation:
We are given that
Diffraction grating=600 lines/mm
d=
Wavelength of light,
l=4.6 m
A.We have to find the distance between the two m=1 bright fringes

For first bright fringe, =1


The distance between two m=1 fringes

Hence, the distance between two m=1 fringes=2.95 m
B.For maximum number of fringes,


Substitute the values


Maximum number of bright fringes on the scree=
Answer:
So... for the element of NITROGEN, you already know that the atomic number tells you the number of electrons. That means there are 7 electrons in a nitrogen atom. Looking at the picture, you can see there are two electrons in shell one and five in shell two.
Answer:
I believe the answer is B.
C, N and O all belong to the same period, in which it's 2nd Period.