<span>There is six horizen.
1. O Horizon - The top, organic layer of soil,
2. A Horizon - The layer called topsoil;
3. E Horizon - This layer is beneath the A Horizon and above the
B Horizon. It is made up mostly of sand.
4. B Horizon - Also called the subsoil - this layer is beneath the E
Horizon and above the C Horizon.
5. C Horizon - it's called regolith: the layer beneath the B Horizon
and above the R Horizon.
6 R Horizon - this is last and the unweathered rock layer that is
beneath all the other layers.</span>
Answer:
The most common products include aerosols, anti-freeze, asbestos, fertilizers, motor oil, paint supplies, photo chemicals, poisons, and solvents, cleaning supplies.
Explanation:
Use homemade cleaners
You can find local retailers to take rechargeable batteries from laptop computers, cordless power tools, cellular and cordless phones, and camcorders at the Rechargeable Battery Recycling Corporation’s website
Answer:
a) t = 0.0185 s = 18.5 ms
b) T = 874.8 N
Explanation:
a)
First we find the seed of wave:
v = fλ
where,
v = speed of wave
f = frequency = 810 Hz
λ = wavelength = 0.4 m
Therefore,
v = (810 Hz)(0.4 m)
v = 324 m/s
Now,
v = L/t
where,
L = length of wire = 6 m
t = time taken by wave to travel length of wire
Therefore,
324 m/s = 6 m/t
t = (6 m)/(324 m/s)
<u>t = 0.0185 s = 18.5 ms</u>
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b)
From the formula of fundamental frquency, we know that:
Fundamental Frequency = v/2L = (1/2L)(√T/μ)
v = √(T/μ)
where,
T = tension in string
μ = linear mass density of wire = m/L = 0.05 kg/6 m = 8.33 x 10⁻³ k gm⁻¹
Therefore,
324 m/s = √(T/8.33 x 10⁻³ k gm⁻¹)
(324 m/s)² = T/8.33 x 10⁻³ k gm⁻¹
<u>T = 874.8 N</u>
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
