Answer:
She will use one 2 as a subscript.
Explanation:
Nitrogen is a chemical element with the symbol "N" and has an atomic number of 7. Thus, it is found in group (5) of the periodic table and as such it has 5 electrons in its outermost shell. Therefore, nitrogen has two (5) valence electrons.
On the other hand, oxygen has an atomic number of 8 and with the symbol "O."
When nitrogen and oxygen react chemically, they produce a compound known as nitrogen dioxide
In this scenario, Jenny is studying a compound that has two oxygen atoms and one nitrogen atom. Therefore, the statement which describes the subscripts she will use to write the chemical formula is, she will use one 2 as a subscript.

Where: 2 represents the subscript of oxygen.
Answer:
a) 
b) 
c) 
d) 
Explanation:
<u>Given equation of pressure variation:</u>
![\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20%281.78%5C%20Pa%29%5C%20sin%5C%20%5B%280.888%5C%20m%5E%7B-1%7D%29x-%28500%5C%20s%5E%7B-1%7D%29t%5D)
We have the standard equation of periodic oscillations:

<em>By comparing, we deduce:</em>
(a)
amplitude:

(b)
angular frequency:


∴Frequency of oscillations:


(c)
wavelength is given by:



(d)
Speed of the wave is gives by:



Answer:
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
Explanation:
Given data
initial circumference = 165 cm
rate = 12.0 cm/s
magnitude = 0.500 T
tome = 9 sec
to find out
emf induced and direction
solution
we know emf in loop is - d∅/dt ........1
here ∅ = ( BAcosθ)
so we say angle is zero degree and magnetic filed is uniform here so that
emf = - d ( BAcos0) /dt
emf = - B dA /dt ..............2
so area will be
dA/dt = d(πr²) / dt
dA/dt = 2πr dr/dt
we know 2πr = c,
r = c/2π = 165 / 2π
r = 26.27 cm
c is circumference so from equation 2
emf = - B 2πr dr/dt ................3
and
here we find rate of change of radius that is
dr/dt = 12/2π = 1.91
cm/s
so when 9.0s have passed that radius of coil = 26.27 - 191 (9)
radius = 9.08
cm
so now from equation 3 we find emf
emf = - (0.500 ) 2π(9.08
) 1.91 
emf = - 0.005445
and magnitude of emf = 0.005445 V
so
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise
It is trajectory acceleration. A friction track is a device to study motion in low friction environments, I believe. Does this help?
Answer: 
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q = Heat absorbed=
Joules
m= mass of copper = 454 g
c = specific heat capacity = 
Initial temperature of the copper =
= 23.0°C
Final temperature of the water =
= ?
Change in temperature ,
Putting in the values, we get:


The final temperature of copper will be 