Part 1)
here we know that supply took 10 s to reach the ground
so here we will have
Part 2)
Here all the supply covered horizontal distance of 650 m in 10 s interval of time
so here we can say
Answer:
F = - 3.53 10⁵ N
Explanation:
This problem must be solved using the relationship between momentum and the amount of movement.
I = F t = Δp
To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio
v = d / t
t = d / v
Reduce SI system
m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg
d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m
Let's calculate
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
With this value we use the momentum and momentum relationship
F t = m v - m v₀
As the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵
F = - 3.53 10⁵ N
The negative sign indicates that the force is exerted against the bullet
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
