Answer to Question #90205 in Physics for Paris Phoebe

Question #90205
Find the magnitude and direction of the following vector represented by the following pairs of components
a) Ax=8.60cm. Ay=5.20cm
b) Ax=9.70m. Ay=2.45m
c) Ax= 7.75km. Ay=2.70km
1
Expert's answer
2019-05-27T11:09:37-0400

a) We can find the magnitude of the vector A\overrightarrow{A} from the formula:


A=A=Ax2+Ay2=(8.60cm)2+(5.20cm)2=10cm.A = \lvert \overrightarrow{A} \rvert = \sqrt{A_x^2 + A_y^2} = \sqrt{(8.60 cm)^2 + (5.20 cm)^2} = 10 cm.

We can find the angle between the vector A\overrightarrow{A} and the +x+x-axis from the formula:


θ=arctan(AyAx)=arctan(5.20cm8.60cm)=31.\theta = arctan(\dfrac{A_y}{A_x}) = arctan(\dfrac{5.20 cm}{8.60 cm}) = 31^{\circ}.

b) We can find the magnitude of the vector A\overrightarrow{A} from the formula:


A=A=Ax2+Ay2=(9.70m)2+(2.45m)2=10m.A = \lvert \overrightarrow{A} \rvert = \sqrt{A_x^2 + A_y^2} = \sqrt{(9.70 m)^2 + (2.45 m)^2} = 10 m.

We can find the angle between the vector A\overrightarrow{A} and the +x+x-axis from the formula:


θ=arctan(AyAx)=arctan(2.45m9.70m)=14.\theta = arctan(\dfrac{A_y}{A_x}) = arctan(\dfrac{2.45 m}{9.70 m}) = 14^{\circ}.

c) We can find the magnitude of the vector A\overrightarrow{A} from the formula:


A=A=Ax2+Ay2=(7.75km)2+(2.70km)2=8.21km.A = \lvert \overrightarrow{A} \rvert = \sqrt{A_x^2 + A_y^2} = \sqrt{(7.75 km)^2 + (2.70 km)^2} = 8.21 km.

We can find the angle between the vector A\overrightarrow{A} and the +x+x-axis from the formula:


θ=arctan(AyAx)=arctan(2.70km7.75km)=19.21.\theta = arctan(\dfrac{A_y}{A_x}) = arctan(\dfrac{2.70 km}{7.75 km}) = 19.21^{\circ}.

Answer:

a) A=10cmA = 10 cm, θ=31.\theta = 31^{\circ}.

b) A=10mA = 10 m, θ=14.\theta = 14^{\circ}.

c) A=8.21kmA = 8.21 km, θ=19.21.\theta = 19.21^{\circ}.


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