Answer:
v (minimum speed) = 2.90 m/sec.
Maximum value of speed will occur at lowest point of vertical circle.
Explanation:
a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?
Using the force balance expression at the top of the circle,
Gravitational Force + Tension force = Centrifugal force
Given that : T = 0
R = length of string = 0.86 m
mass of the spinning rock = 0.75 kg
v (minimum speed) = 2.90 m/sec.
b) what is the maximum speed the rock can have so that the string does not break?
Here the force balance at bottom of circle is represented by the illustration:
Given that:
maximum tension T = 45 N
maximum speed v = ??
mass m = 0.75 kg
∴
c)
At what point in the vertical circle does this maximum value occur?
Maximum value of speed will occur at lowest point of vertical circle.
This is so because at the lowest point; the tension in string will be maximum.
if currents go in opposite directions, wires repel
Answer:
a = 52s²
Explanation:
<u>How to find acceleration</u>
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
<u>Solve</u>
We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)
We first need to solve the velocity equation for time (t):
v = u + at
v - u = at
(v - u)/a = t
Plugging in the known values we get,
t = (v - u)/a
t = (16 m/s - 120 m/s) -2/s2
t = -104 m/s / -2 m/s2
t = 52 s
Work is done when a force is applied to an object moves that object. the work is calculated by multiplying the force by the amount of movement of an object
Answer:
v₂ = 7/ (0.5)= 14 m/s
Explanation:
Flow rate of the fluid
Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.
The formula for calculated the flow rate is:
Q= v*A Formula (1)
Where :
Q is the Flow rate (m³/s)
A is the cross sectional area of a section of the pipe (m²)
v is the speed of the fluid in that section (m/s)
Equation of continuity
The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:
Q₁= Q₂
Data
A₁ = 2m² : cross sectional area 1
v₁ = 3.5 m/s : fluid speed through A₁
A₂ = 0.5 m² : cross sectional area 2
Calculation of the fluid speed through A₂
We aply the equation of continuity:
Q₁= Q₂
We aply the equation of Formula (1):
v₁*A₁= v₂*A₂
We replace data
(3.5)*(2)= v₂*(0.5)
7 = v₂*(0.5)
v₂ = 7/ (0.5)
v₂ = 14 m/s