Answer:
Explanation:
Given
Length of plank is 1.6 m
Force is applied on the left side of plank
Force is applied 43 cm from the left end O.
Mass of the plank is
for equilibrium
Net torque must be zero. Taking torque about left side of the plank
Net vertical force must be zero on the plank
Answer:
Explanation:
From the given information:
We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then
The volume charge distribution relates to the radial direction at r = R
∴
To find the constant k, we examine the total charge Q which is:
∴
Thus;
Hence, from equation (1), if k =
To verify the units:
↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q
since
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
A motorboat accelerates uniformly from a velocity of 6.5m/s
to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2
to the east ,
how far did it travel during the accelration? Give your
answer in units of kilometers per hour/sec. To find the acceleration of the car
we have to
<span>
1. First determine
the suitable formula for this word problem.
Which is a. A=vf-vi/t</span>
which will be
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s
Solution:
<span>
x = v0t + ½at2</span>
<span>x = <span>16.645375 m </span></span>
An example of a a predator in the Ethiopian Highlands ecosystem is the
Wolf.The wolf is a carnivore which acts as a predator to other smaller
animals in the ecosystem.
<h3>What is a Prey?</h3>
Preys are mostly smaller animals in which the predators feed on for food.
They are usually herbivores and primary consumers in the ecosystem.
Examples of Preys include:
Read more about Feeding relationship here brainly.com/question/9852437