Answer:

Explanation:
given,
side slope = 1 : 1
hydraulic depth(y) = 5 ft
bottom width (b)= 8 ft
x = 1
Q = 2,312 ft³/s
n = 0.013
slope of channel = ?



R = 4.69 m
using manning's equation



S = 0.406

2.71 m/s fast Hans is moving after the collision.
<u>Explanation</u>:
Given that,
Mass of Jeremy is 120 kg (
)
Speed of Jeremy is 3 m/s (
)
Speed of Jeremy after collision is (
) -2.5 m/s
Mass of Hans is 140 kg (
)
Speed of Hans is -2 m/s (
)
Speed of Hans after collision is (
)
Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is
= 
Substitute the given values,
= 120 × 3 + 140 × (-2)
= 360 + (-280)
= 80 kg m/s
Linear momentum after the collision of Jeremy and Hans is
= 
= 120 × (-2.5) + 140 × 
= -300 + 140 × 
We know that conservation of liner momentum,
Linear momentum before the collision = Linear momentum after the collision
80 = -300 + 140 × 
80 + 300 = 140 × 
380 = 140 × 
380/140= 
= 2.71 m/s
2.71 m/s fast Hans is moving after the collision.
Answer:

Explanation:
We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.
We know that frequency of simple pendulum follows that 
Now, the speed of the father will be
while for the child the speed will be 
The ratio of the father’s speed to the child’s speed will be

Answer:
220 A
Explanation:
The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.
So, F = BI₁L
F = (μ₀I₂/2πd)I₁L
F = μ₀I₁I₂L/2πd
Given that the current in the rods are the same, I₁ = I₂ = I
So,
F = μ₀I²L/2πd
Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²
So, F = W
μ₀I²L/2πd = mg
making I subject of the formula, we have
I² = 2πdmg/μ₀L
I = √(2πdmg/μ₀L)
substituting the values of the variables into the equation, we have
I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])
I = √(0.0049 × 10⁷kgm²/s²H)
I = √(0.049 × 10⁶kgm²/s²H)
I = 0.22 × 10³ A
I = 220 A
Answer:
first object final velocity =2m/s
second = 12m/s
Explanation:
i hope this will help you,..,