Answer:
B
Explanation:
Given:-
- The charge of the test particle q = 3.0 * 10^-9 C
- The force exerted by the metal sphere F = 6.0 * 10^-5 N
Find:-
The magnitude and direction of the electric field
strength at this location?
Solution:-
- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:
F = E*q
- Using the data given we can determine E:
E = F / q
E = (6.0 * 10^-5) / (3.0 * 10^-9)
E = 20,000 N/C
- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.
Answer:true
Explanation:
Displacement is the vector representation of a change in position. It is path independent and is equivalent to the straight line distance between the start and end locations. Distance is a scalar quantity that reflects the path traveled.
F = m . g = 76.5 x 9..8 = 749.7
Net Force = 3225 - 749.7 = 2475.3
F = m.a
2475.3 = 76.5 a
a = 32.35
V = at + v1
V = at + 0
V = 32.35 x 0.15
V = 4.8525
Hope this helps
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
