Answer:
The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.
Explanation:
F= 280 N
m= 80 kg
α= 12º
μ= 0.15
d= 100m
g= 9,8 m/s²
N= m*g*sin(α)
N= 163 Newtons
Fr= μ * N
Fr= 24.45 Newtons
∑F= m*a
a= (280N - 24.5N) / 80kg
a= 3.19 m/s²
d= a * t² / 2
t=√(2*d/a)
t= 7.91 sec
V= a* t
V= 3.19 m/s² * 7.91 s
V= 25.23 m/s
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:

where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)

A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr
Answer:
a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J
Explanation:
a) The inductance is a solenoid this given carrier
L =
The magnetic field inside the solenoid is
B = μ₀
hence the magnetic flux
Ф_B = B. A = μ₀
we substitute in the expression of inductance
L = N² μ₀ A /l
let's find the area of each turn
A = π r²
A = π 0.02²
A = 1.2566 10⁻³ m²
let's calculate
L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3
L = 3.29 10⁻⁴ H
b) The stored energy is
U = ½ L i²
let's calculate
U = ½ 3.29 10⁻⁴ 18²
U = 5.33 10⁻² J