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tatyana61 [14]
2 years ago
6

A person walks 9 meters to the right and 8 meters to the left what's the distance and displacement​

Physics
2 answers:
nordsb [41]2 years ago
5 0
The distance is 17 and the displacement is 1
saw5 [17]2 years ago
4 0

Answer:

walked: 17 meters

displacement: 1

Explanation:

so you add 9+8 and the displacement is 1 b/c when you moved back those 8 meters you were still 1 meter away from where you where originally standing

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Answer:

twelve facesFrom left to right the solids are tetrahedron (four sides), cube (six sides), octahedron (eight faces), dodecahedron (twelve faces), and icosahedron (twenty faces).

Explanation:

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Substances released into the air are known as
irakobra [83]

Your answer is Emissions

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Un acróbata de 60.0 kg está unido a un cordón de bungee con un resorte de 10.0 m de longitud . Salta de un puente que abarca un
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Answer:

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8 0
2 years ago
What is the identity of a sample that has a volume of 92.5 cm3 and a mass of 249.8 g?
denpristay [2]
Here, We know, Density = Mass / Volume
Here, mass = 249.8 g
volume = 92.5 cm³

Substitute their values, 
d = 249.8 / 92.5
d = 2.7 g/cm³

In short, Your Answer would be "Aluminum" 

Hope this helps!
3 0
3 years ago
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In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of
Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, a_{Net} = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, v_y = v × sin(θ)

∴ v_y = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - a_{Net} × t

∴ t = (v₁ - v₂)/a_{Net}  = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2·a_{Net}·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0
2 years ago
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