Answer:
A) At point 1, local acceleration = 0.5 m/s²
At point 2, local acceleration = 1.0 m/s²
B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Explanation:
Local acceleration at those points is the instantaneous acceleration at those points and it is given as
a = dv/dt
At point 1, v₁ = 0.5 t
a₁ =dv₁/dt = 0.5 m/s²
At point 2, v₂ = 1.0 t
a₂ = dv₂/dt = 1.0 m/s²
b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time
Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t
Time = t
Average acceleration = 0.5t/t = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Answer:
1) The net electric field at any location inside a block of copper is zero if the copper block is in equilibrium.
2) In equilibrium, there is no net flow of mobile charged particles inside a conductor.
3) If the net electric field at a particular location inside a piece of metal is not zero, the metal is not in equilibrium.
Explanation:
1) and 3) A block of copper is a conductor. The charged particles on a conductor in equilibrium are at rest, so the intensity of the electric field at all interior points of the conductor is zero, otherwise, the charges would move resulting in an electric current.
2) The charged particles on a conductor in equilibrium are at rest.
Answer:
96%
Explanation
Let A the total area of the galaxy, is modeled as a disc:
A = πR^2 = π (25 kpc)^2
And let a be the area that astronomers are able to see:
a = πr^2 = π(5 kpc)^2
The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:
P = 100 a/A = (5/25)^2 = 100/25 = 4%
Therefore, the percentage of the galaxy not included, i.e. not seen is:
(100-4)% = 96%
Answer: car B has travelled 4times as far as Car A
d=vi*t+1/2at^2
No initial velocity so equation becomes;
d=1/2at^2 and the acceleration is the same between both only time is different;
Car A d=1/2a(1)^2
Car B d=1/2a(2)^2
Car A d= 1^2=1
Car B d= 2^2=4
Car B d=4*Car A
So car B has travelled 4 times as far as car A
