Answer:
That an item is neither moving nor staying still in a position that is building up energy.
Explanation:
This seems like an incomplete question..
Seven
The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer
Eight
They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.
The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read. the difference is 360.
I really do object to the wording, but what can I do?
Nine
Nine is the same thing as 8.
Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408
The actual distance is 400 + 80 = 480
The difference is the answer = 480 - 408 = 72 <<<< Answer
Ten
This is just the displacement magnitude.
dis = sqrt(30^2 + 80^2)
dis = sqrt(900 + 6400)
dis = sqrt(7300)
dis = 85.44 <<<< Answer D
Twelve
Vi = 2.15*Sin(30) = 1.075 m/s
vf = 0
a = - 9.81
t = ?
<u>Formula</u>
a = (vf - vi)/t
<u>Solve</u>
-9.81 = (0 - 1.075)/t
- 9.81 * t = -1.075
t = 0.11 seconds
Thirteen
I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.
Edit
That is a surprise! Really quickly
d = 3.2 m
a = - 9.82
vf = 0
vi = ?
vf^2 = vi^2 - 2*a*d
0 = vi^2 - 2*9.81*3.2
vi = sqrt(19.62*3.2)
vi = 8.0 m/s But that is the vertical component of the speed
v = vi/sin(25)
v = 8.0/sin(25) = 11
Answer:
serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
1 / Ceq = 1,147
Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
Dv = Q / Ceq
Q = DV Ceq
Q = 14 0.678 10⁻⁶
Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
ΔV = Q / C5
ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
ΔV = 1,725 V
d) The energy stores is
U = ½ C V²
U = ½ 0.678 10-6 14²
U = 66.4 10⁻⁶ J
e) Parallel system
Ceq = C1 + C2 + C3 + C4 + C5
Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
Q5 = C5 V
Q5 = 5.5 10⁻⁶ 14
Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
U = ½ C V²
U = ½ 22.7 10-6 14²
U = 2.2 10⁻³ J
It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
