A wave is a disturbance that transfers energy through medium from one place to another. Do the particles in the medium travel w
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The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864
Information about the problem:
- Space long= 84 ft
- Space wide= 42 ft
- Space deep= 9 ft
- Cost by yard3 = $39/yd3
- Total cost= ?
To solve this problem, we have to state the equation using the information of the problem:
Calculating the volume of the total space:
space volume = space long * space wide * space deep
space volume = 84 ft * 42 ft * 9 ft
space volume = 31752 ft3
By converting the volume from ft3 to yd3, we have:
31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3
Calculating the cost of excavating the volume space:
Total cost = space volume * cost by yard3
Total cost = 1176 yd3 * $39/yd3
Total cost = $45864
<h3>What is volume?</h3>It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.
Learn more about volume at: brainly.com/question/12628341
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct