Answer: The younger elliptical and lenticular galaxies had results similar to spiral galaxies like the Milky Way. The researchers found that the older galaxies have a larger fraction of low-mass stars than their younger counterparts.
Explanation:
Answer : The maximum concentration of silver ion is 
Solution : Given,
for AgBr = 
Concentration of NaBr solution = 0.1 m
The equilibrium reaction for NaBr solution is,

The concentration of NaBr solution is 0.1 m that means,
![[Na^+]=[Br^-]=0.1m](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D%5BBr%5E-%5D%3D0.1m)
The equilibrium reaction for AgBr is,

At equilibrium s s
The expression for solubility product constant for AgBr is,
![K_{sp}=[Ag^+][Br^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D)
The concentration of
= s
The concentration of
= 0.1 + s
Now put all the given values in
expression, we get

By rearranging the terms, we get the value of 's'

Therefore, the maximum concentration of silver ion is
.
a). for velocity, you must have a number, a unit, and a direction.
Yes. This one isn't bad. The 'number' and the 'unit' are the speed.
b). the si units for velocity are miles per hour.
No. That's silly.
'miles' is not an SI unit, and 'miles per hour'
is only a speed, not a velocity.
c). the symbol for velocity is .
You can use any symbol you want for velocity, as long as
you make its meaning very clear, so that everybody knows
what symbol you're using for velocity.
But this choice-c is still wrong, because either it's incomplete,
or else it's using 'space' for velocity, which is a very poor symbol.
d). to calculate velocity, divide the displacement by time.
Yes, that's OK, but you have to remember that the displacement
has a direction, and so does the velocity.
Answer:
50 meters
Explanation:
Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

Since the car starts at rest, you can write the following equation:

Now that you have the acceleration, you can do this:

Once again, there is no initial velocity:

Hope this helps!
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q = 
=
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is
.