I will assume you are asking what the initial acceleration of the sphere is since the information provided seems to indicate that.
First we need to know Newton's Law
F=ma.
We know the mass of the sphere and we want a so we solve to get
a=F/m.
Now we need the force on the charged sphere. This is given by the electric field, E and the charge, Q. The relationship is F=Q×E. (Recall that the electric field units can be expressed in Newtons/Coulomb).
Now the electric field above a large (~infinite) sheet of charge with a known charge density σ, is given by
E = σ/(2ε0)
Plug in your values of σ, to get E, then the sphere charge Q to get F, the the mass into a = F/m to get the acceleration
<span>s=2.7 centimeters = 0.027 meters
t=3.9 milliseconds = 0.0039 seconds
s=(1/2)a*t^2
so
a=(2.7*2)/(0.0039)^2
= 355,029.58 m/s^2
a=355,029.58 m/s^2 = 355.02958 km / s^2</span>
V2 = 4.4579 L
Since pressure is constant, use Charle’s law.
Charles's law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.
V(olume) 1 = V(olume) 2
————— —————
T(emperature) 1 T(emperature)2
4.00 L = V2
———- ———
297 K 331 K
Cross multiply
(4.00 L x 331 K) = (297 K x V2)
Simplify
1324 L/K = 297 K x V2
Isolate V2 by dividing out 297 K
1324 L/K = V2
————
297 K
(This cancels out the kelvin and leaves you with Liters as the volume measure)
V2 = 4.4579 L
Round to significant digits if required
For the given question above, I think there is an associated choice of answer for it. However, the answer for this is London Dispersion Forces. <span>Dipole-dipole forces and hydrogen bonding are much stronger, leading to higher melting and boiling points.</span>
Hello there!
Essentially, a control variable is what is kept the same throughout the experiment, and it is not of primary concern in the experimental outcome. Any change in a control variable in an experiment would invalidate the correlation of dependent variables (DV) to the independent variable (IV), thus skewing the results.
