Answer:


Explanation:
Given that
Q= 5 L/min
1 L = 10⁻³ m³/s
1 min = 60 s
Q=0.083 x 10⁻³ m³/s
d= 6 μm
v= 1 mm/s
So the discharge flow through one tube
q = A v


A=2.8 x 10⁻¹¹ m²
v= 1 x 10⁻³ m/s
q= 2.8 x 10⁻¹⁴ m³/s
Lets take total number of tube is n
Q= n q
n=Q/q


Surface area A
A= π d L


Answer:
When the elevator is accelerating downward, the person feels lighter due to the downward normal force being less than the person's weight.
Explanation:
A person riding in an elevator subjected to a series of unbalanced forces depending on the direction the elevator is travelling.
Two forces are acting on the person; the force of gravity and the upward normal force from the elevator.
When the elevator is going upwards with acceleration a, the person feels heavier than his normal weight, due to the upward normal force being greater than the person's weight. N = mg + ma
When the elevator is moving downwards with acceleration a, the person feels lighter due to the downward normal force being less than the person's weight. N = mg - ma
However, when the elevator is moving up or down at constant velocity ie. acceleration a = 0, the person experience a normal force equal to weight. N = mg
When the elevator is moving downwards with acceleration a = g, the person experiences weightlessness. N = (mg - mg) = 0
Answer:
i = 4.9 A
Explanation:
The expression for the magnetic force in a wire carrying a current is
F = i L x B
bold letters indicate vectors.
The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition
F - W = 0
i L B = m g
They indicate the linear density of the cable λ = 0.2 kg / m
λ = m / L
m = λ L
we substitute
i B = λ g
i = 
let's calculate
i = 0.2 9.8 / 0.4
i = 4.9 A
from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"
here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Answer:
(A) 26 m/s
(B) 32.4 m
(C) v = 15.4 m/s
Explanation:
initial speed (u) = 6.4 m/s
acceleration due to gravity (a) = 9.9 m/s^[2}
time (t) = 2 s
(A) What is its speed after falling for 2.00s?
from the equation of motion v = u + at we can get the speed
v = 6.4 + (9.8 x 2) = 26 m/s
(B) How far does it fall in 2.00s?
from the equation of motion
we can get the distance covered
s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)
s = 12.8 + 19.6 = 32.4 m
c) What is the magnitude of its velocity after falling 10.0m?
from the equation of motion below we can get the velocity

v = 15.4 m/s