
According to above question ~
Let's find the charge (q) by using formula ~
Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds
Answer:
I(x) = 1444×k ×
I(y) = 1444×k ×
I(o) = 3888×k ×
Explanation:
Given data
function = x^2 + y^2 ≤ 36
function = x^2 + y^2 ≤ 6^2
to find out
the moments of inertia Ix, Iy, Io
solution
first we consider the polar coordinate (a,θ)
and polar is directly proportional to a²
so p = k × a²
so that
x = a cosθ
y = a sinθ
dA = adθda
so
I(x) = ∫y²pdA
take limit 0 to 6 for a and o to
for θ
I(x) =
y²p dA
I(x) =
(a sinθ)²(k × a²) adθda
I(x) = k
da ×
(sin²θ)dθ
I(x) = k
da ×
(1-cos2θ)/2 dθ
I(x) = k
×
I(x) = k ×
× (
I(x) = k ×
×
I(x) = 1444×k ×
.....................1
and we can say I(x) = I(y) by the symmetry rule
and here I(o) will be I(x) + I(y) i.e
I(o) = 2 × 1444×k ×
I(o) = 3888×k ×
......................2
This is the same question as the one previously but with more details, so I will just use my previous answer.
1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.
So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
Answer:
The time taken to stop the box equals 1.33 seconds.
Explanation:
Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

Given mass of box = 5.0 kg
Frictional force = 30 N
thus

Now to find the time that the box requires to stop can be calculated by first equation of kinematics
The box will stop when it's final velocity becomes zero

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.