When we jump off a boat towards the shore the boat moves back why
1 answer:
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Answer:
Explanation:
Given
Length of plank is 1.6 m
Force is applied on the left side of plank
Force is applied 43 cm from the left end O.
Mass of the plank is
for equilibrium
Net torque must be zero. Taking torque about left side of the plank
Net vertical force must be zero on the plank
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