A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and directio
n of the resultant displacement, using the method of components. In a vector-addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components
1 answer:
Answer: magnitude of resultant = 2.811km, direction of resultant = 209.6°
Explanation: The vector ( in this case displacement) that lies on the y axis is 3.25km north and 1.50km due south.
Their resultant is gotten below
Magnitude of resultant = 3.25 - 1.50 = 1.75km
Direction of resultant = north (direction of the bigger vector)
2.20km is the only vector acting on the x axis due east, combining this vector with the resultant of the vectors above, we realize that 2.2km west is perpendicular to 1.75km due north. Since 1.75km us due north, it implies that it is the vector on the positive y axis (vy) and 2.20km due west implies that it is the vector on the negative x axis(vx), thus their resultant is gotten using phythagoras theorem.
R = √ vx² + vy²
R = √ 2.20² + 1.75²
R = √ 7.9025
R= 2.1811km.
θ = tan^-1 (vy/vx)
θ = tan ^-1 (1.25/2.20)
θ = tan ^-1 (0.5618)
θ = 29.6°.
The direction of the vector is south west which implies the third quadrant of the trigonometric quadrant which implies 180 + θ
Thus the direction of the vector is 180 + 29.60 = 209.6°
From the picture attached to this answer we can see roughly that the magnitude of resultant is longer than it component and the vector is placed on the 3rd quadrant which verifies our quantitative claim.
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Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) =
V(r) - V(0) = ₀
V(r) = ₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) =
V(r) - V(0) =
V(r) =
V(r) =
V(r)
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V