What can you infer about the changes in the environment over time at the mystery fossil dig site? Record your inferences in the
1 answer:
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Answer:
a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º
This is 8º north west
Explanation:
This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately
To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.
First vector A = 2 to 20º north west
Measured from the positive x axis is θ = 180 -20 = 160º
We use trigonometry to find the components
Cos 20 = Aₓ / A
sin 20 = / A
Aₓ = A cos 160 = 2 cos 160
= A sin160 = 2 sin160
Aₓ = -1,879 mi
= 0.684 mi
Second vector B = 4 mi 10º west of the south
Angle θ = 270 - 10 = 260º
cos 2600 = Bₓ / B
sin 260 = / B
Bₓ = B cos 260
= B sin 260
Bₓ = 4 cos 260
= 4 sin 260
Bₓ = -0.6946mi
= - 3,939 mi
Third vector C = 3 mi to 15 north east
cos 15 = Cₓ / C
sin15 = / C
Cₓ = C cos 15
= C sin15
Cₓ = 3 cos 15
= 3 sin 15
Cₓ = 2,898 mi
= 0.7765 mi
Now we can find the final position of the person
X = Aₓ + Bₓ + Cₓ
X = -1.879 -0.6949 + 2.898
X = 0.3241 mi
Y = +
+
Y = 0.684 - 3.939 +0.7765
Y = -2.4785 mi
a) We use Pythagoras' theorem
R = √ (x2 + y2)
R = √ (0.3241 2 + (-2.4785) 2)
R = 2.4996 mi
R = 2.5 mi
b) let's use trigonometry
Tan θ = y / x
Tanθ = -2.4785 / 0.3241
θ = tan⁻¹ (-7,647)
θ = -82
Measured from the positive side of the x axis is Te = 360 - 82 = 278º
(90-82) south east
To return to your case you must walk in the opposite direction or Te = 98º
This is 8º north west
Answer:
= ( ρ_fluid g A) y
Explanation:
This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force
for the first part, let's write Newton's equilibrium equation
B₀ - W = 0
B₀ = W
ρ_fluid g V_fluid = W
the volume of the fluid is the area of the cube times the height it is submerged
V_fluid = A y
For the second part, the body introduces a quantity and below this equilibrium point, the equation is
B - W = m a
ρ_fluid g A (y₀ + y) - W = m a
ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a
ρ_fluid g A y + (B₀-W) = ma
the part in parentheses is zero since it is the force when it is in equilibrium
ρ_fluid g A y = m a
this equation the net force is
= ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position