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Vector E₁ is an electric field at the origin with magnitude of 3.75 × 10³ N/C directed 35° below the +x axis, while vector E2 is the electric field at the origin with magnitude 5.0 x 105 N/C directed 20° to the left of the +y axis. a) Do a good-sized (about ½ page) well-labeled sketch of this, b) show the vector addition graphically on your sketch to eyeball accuracy, and c) find the magnitude and direction of the net force vector Etot E1 E₂. -

Vector E₁ is an electric field at the origin with magnitude of 3.75 × 10³ N/C directed 35° below the +x axis, while vector E2 is the electric field at the origin with magnitude 5.0 x 105 N/C directed 20° to the left of the +y axis. a) Do a good-sized (about ½ page) well-labeled sketch of this, b) show the vector addition graphically on your sketch to eyeball accuracy, and c) find the magnitude and direction of the net force vector Etot E1 E₂. -

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Hello, i need help understanding how do you find the sin and cosine of E1 x and y along with E2 x and y. This problem is complete that im showing here please explain with this problem how you got the sin and cosine to start the problem for vector addition!!!

Vector E₁ is an electric field at the origin with magnitude of 3.75 × 10³ N/C directed 35° below the +x axis, while vector E2 is the electric field at the origin with magnitude 5.0 × 10³ N/C directed 20° to the left of the +y axis. a) Do a good-sized (about ½ page) well-labeled sketch of this, b) show the vector addition graphically on your sketch to eyeball accuracy, and c) find the magnitude and direction of the net force vector Eto=E1 E2. y 2₂ 350 16
Transcribed Image Text:Vector E₁ is an electric field at the origin with magnitude of 3.75 × 10³ N/C directed 35° below the +x axis, while vector E2 is the electric field at the origin with magnitude 5.0 × 10³ N/C directed 20° to the left of the +y axis. a) Do a good-sized (about ½ page) well-labeled sketch of this, b) show the vector addition graphically on your sketch to eyeball accuracy, and c) find the magnitude and direction of the net force vector Eto=E1 E2. y 2₂ 350 16
X E₁ +E, C350 = 3.75w35 =+3.07 L E₂-E₂ Ain 200 E tot 40 보 -E, sin 35° =-3.75Ain 35* = -2.15 =-5,0 in 20⁰ + E₂200 = + 5.0 co20⁰ = + 4.70 =-1.71 1.36x105 / +2.55×108 N/C (x10³² NC Ext(1.36x054) + (2.55×10²4) ² = √2.89×10²N/C angle: choose one of a છે ન o tan 8 = Fisty | = ta 2.55.105 Extx 1.36X 105 N/C ⇒0=61.9° 1.36 2.55 $ Exty => 4 = 28.1 °
Transcribed Image Text:X E₁ +E, C350 = 3.75w35 =+3.07 L E₂-E₂ Ain 200 E tot 40 보 -E, sin 35° =-3.75Ain 35* = -2.15 =-5,0 in 20⁰ + E₂200 = + 5.0 co20⁰ = + 4.70 =-1.71 1.36x105 / +2.55×108 N/C (x10³² NC Ext(1.36x054) + (2.55×10²4) ² = √2.89×10²N/C angle: choose one of a છે ન o tan 8 = Fisty | = ta 2.55.105 Extx 1.36X 105 N/C ⇒0=61.9° 1.36 2.55 $ Exty => 4 = 28.1 °
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