Answer:Velocity = 6.325m/s
Directional angle= 18.43°
Explanation:
Using Right angle triangle
Let Velocity of ballon&hawk be VHB represent the height of the triangle.
Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.
Let Velocity of hawk and ground BE VHG represent the hypothesis.
Theta = opp/Adj= VHB/VBG
using pythagorean
VHG= SQRT(VHB^2+VBG^2)
VHG= sqrt(2^2+6^2)
VHG= sqrt(4+36)
VHG= 6.325m/s
Tan theta= 2/6
Tan theta =0.3333
Tan^-1 0.3333=18.43°
Answer:
magnitude: 21.6; direction: 33.7 degrees
Explanation:
When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have
vector: (-3,-2)
Scalar: -6
so the vector multiplied by the scalar will have components
The magnitude is given by Pythagorean's theorem:
and the direction is given by the arctan of the ratio between the y-component and the x-component:
Answer:
1 / f = 1 / i + 1 / o thin lens equation
1 / i = 1 / f - 1 / o = (o - f) / (o * f)
i = o * f / (o - f)
i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm image distance
Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall
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