Answer:
The square plates of a 3000-pF parallel-plate capacitor measure 40 mm by 40 mm and are separated by a dielectric that is 0.29 mm thick and completely fills teh region between the plates.
Explanation:
Answer:
Magnitude = 4.056 m
Direction = 42.3⁰
Explanation:
The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.
The vector 4.40 is directed East. This automatically becomes a horizontal component.
But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.
Resolving the vectors should yield the horizontal and vertical components:
Horizontal components
The first component is 4.40 m
The second one is derived by resolving 3.40 to the horizontal like this 3.40 × - cos 61⁰ = -1.648 m
Adding the horizontal component gives 4.40 m + ( -1.648 m) = 2.752 m
Vertical components
Resolve 3.40 with the angle 61⁰ like this: vertical comp = 3.41 × sin 61
= 2.98 m
The magnitude is given by √[(2.98)²+ (2.752)²] = 4.056 m Ans
The direction us given by tan⁻¹ (2.98/2.752) = 42.3⁰ Ans
Answer: The time required for the impluse passing through each other is approximately 0.18seconds
Explanation:
Given:
Length,L = 50m
M/L = 0.020kg/m
FA = 5.7×10^2N
FB = 2.5×10^2N
The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.
Ca(t) + CB(t) = 50
Where CA and CB are the velocities of the wire A and B
t = 50/ (CA + CB)
But C = Sqrt(FL/M)
Substituting gives:
t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))
t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))
t = 50 / (168.62 + 111.83)
t = 50/280.15
t = 0.18 seconds
Hi, I believe this should belongs to the chemistry section, but nevertheless here is the solution :
Number of Fe atoms: (2) + (2) =4
Number of O atoms. : (3)
Number of H atoms: (3 X 2)=6
Total atoms: 4+3+6=13 atoms
Answer:
Part a)
Part b)
Explanation:
As we know that by parallel axis theorem we will have
Part a)
here we know that for a stick the moment of inertia for an axis passing through its COM is given as
now if we need to find the inertia from its end then we will have
Part b)
here we know that for a cube the moment of inertia for an axis passing through its COM is given as
now if we need to find the inertia about an axis passing through its edge
