So if you take a 39 ft ladder and place its base 15 feet from the base of the wall, when you lean the ladder against the wall, the point where the ladder rests against the wall will be 36 feet above the ground.

Answer: 15 feet
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Hope this helps you understand this problem a little better.

6 0

2 years ago

Help people i really need this

denis-greek [22]

Option “A” is the crest of the wave because it’s the maximum value of upward displacement within a cycle.

5 0

3 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

2 years ago