Answer:
The coefficient of friction in the hall is 0.038
Explanation:
Given;
mass of the Parker, m = 73.2 kg
applied force on the parker, F = 123 N
frictional force, Fs = 27.4 N
the coefficient of friction in the hall = ?
frictional force is given by;
Fs = μN
Where;
μ is the coefficient of friction
N is normal reaction = mg
Fs = μmg
μ = Fs / mg
μ = (27.4) / (73.2 x 9.8)
μ = 0.038
Therefore, the coefficient of friction in the hall is 0.038
Answer:
50 degree.
Explanation:
Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7
The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:
Tan Ø = Ay/Ax
Substitute Ay and Ax into the formula above.
Tan Ø = -4.7 / 5.6
Tan Ø = -0.839
Ø = tan^-1(-0. 839)
Ø = - 40 degree
Therefore, the angle between vector A and B positive direction of x-axis will be
90 - 40 = 50 degree.
Answer:
c.an extinct oceanic hot-spot volcano that has subsided below sea level
Explanation:
- Marine geology defines a guyot as an isolated underwater volcanic mountain with a flat top more than 200m below the surface of the sea.
-The flat top is due to years of wave erosion.
-Guyots can form a chain of seamounts as the ocean plate of the Earth's crust moves slowly over a hot spot that remains stationary beneath the plate.
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