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
You must observe the object twice.
-- Look at it the first time, and make a mark where it is.
-- After some time has passed, look at the object again, and
make another mark at the place where it is.
-- At your convenience, take out your ruler, and measure the
distance between the two marks.
What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point.
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.
it's how much it weighs and how much force is pushing on it like a egg if i drop it the weigh can cause it to break and how much force the gravity is pushing on it.
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
