If a capacitor's dielectric constant is vacuum its dielectric constant is k will be equal to 1.
<u>Explanation:</u>
Relative permittivity of a dielectric substance is referred to as its dielectric constant. Relative permittivity/dielectric constant k is a dimensionless quantity that is the ratio of absolute permittivity and vacuum permittivity.
It is given by the expression
k=k=ε /ε0
where ε denotes absolute permittivity and ε0 denotes permittivity of vacuuum.
Absolute permittivity ε of vacuum= ε0
therefore k= ε0/ ε0=1
dielectric constant of vacuum is 1 .
Momentum = mass • velocity
M = m • v
Speed is a scalar quantity
Velocity is a vector quantity
That is truly the only difference
Impulse = (force) x (length of time the force lasts)
I see where you doodled (60)(40) over on the side, and you'll be delighted
to know that you're on the right track !
Here's the mind-blower, which I'll bet you never thought of:
On a force-time graph, impulse (also change in momentum)
is just the <em>area that's added under the graph during some time</em> !
From zero to 60, the impulse is just the area of that right triangle
under the graph. The base of the triangle is 60 seconds. The
height of the triangle is 40N . The area of the triangle is not
the whole (base x height), but only <em><u>1/2 </u></em>(base x height).
1/2 (base x height) = 1/2 (60s x 40N) = <u>1,200 newton-seconds</u>
<u>That's</u> the impulse during the first 60 seconds. It's also the change in
the car's momentum during the first 60 seconds.
Momentum = (mass) x (speed)
If the car wasn't moving at all when the graph began, then its momentum is 1,200 newton-sec after 60 seconds. Through the convenience of the SI system of units, 1,200 newton-sec is exactly the same thing as 1,200 kg-m/s . The car's mass is 3 kg, so after 60 sec, you can write
Momentum = M x V = (3 kg) x (speed) = 1,200 kg-m/s
and the car's speed falls right out of that.
From 60to 120 sec, the change in momentum is the added area of that
extra right triangle on top ... it's 60sec wide and only 20N high. Calculate
its area, that's the additional impulse in the 2nd minute, which is also the
increase in momentum, and that'll give you the change in speed.
Answer:
Q1_new = 515.68 µC
Q2_new = 246.82 µC
Explanation:
Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.
Then it is only necessary to calculate the charge on each capacitor:
Q1 = 5.85 µF * 250 V = 1462.5 µC
Q2 = 2.8 µF * 250 V = 700 µC
Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.
When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:
1462.5 µC - 700 µC = 762.5 µC
Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.
This 762.5 µC will be divided proportionately between the two capacitors.
Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC
Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC
Answer:
please read the answer below
Explanation:
The angular momentum is given by
By taking into account the angles between the vectors r and v in each case we obtain:
a)
v=(2,0)
r=(0,1)
angle = 90°
b)
r=(0,-1)
angle = 90°
c)
r=(1,0)
angle = 0°
r and v are parallel
L = 0kgm/s
d)
r=(-1,0)
angle = 180°
r and v are parallel
L = 0kgm/s
e)
r=(1,1)
angle = 45°
f)
r=(-1,1)
angle = 45°
the same as e):
L = 5kgm/s
g)
r=(-1,-1)
angle = 135°
h)
r=(1,-1)
angle = 135°
the same as g):
L = 5kgm/s
hope this helps!!
