The flow in systems besides the flow
Answer:
Part a)

Part b)

Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Explanation:
Part a)
As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force
So here we know
buoyancy force is given as



Now we know that the weight of the cylinder is given as

now we have


Part b)
When the same cylinder is floating in other liquid then we will have

so we have


Part c)
Since we know that the base area will remain same always
so here the length and width of the object is not necessary to obtain the above data in such type of questions
Answer:
Answered
Explanation:
A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.
W= FS cosθ
θ= 90 ⇒cos90 = 0 ⇒W= 0
B) work done by tension
W= Tcosθ×S= 5cos30×2.30= 10J
C) Work done by friction force
W= f×s=1×2.30= 2.30 J
D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.
E) The net work done= Work done by tension in the rope - frictional work
=10-2.30= 7.7 J
Answer:
a) E = -4 10² N / C
, b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
-
= m a
Axis y
- W = 0
Initially the system is in equilibrium, so zero acceleration
Fe =
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ =
/ T
T_{y} = cos θ
= T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
= q E
E =
/ q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18
⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
= m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes