An electron moves along the z-axis with vz=4.5×10^7m/s. As it passes the origin, what are the strength and direction of the magn
1 answer:
You might be interested in
Answer:
v = 127.66 m / s θ’= 337.59
Explanation:
For this exercise we must use the speed composition of the drone.
The first speed is 63 m / s in direction 29, three minutes later the speed reaches 22 m / s and direction 233, finally it returns to the launch point with 98 m / s in direction 321, in the attachment you can see a diagram of these speeds.
To find the resulting average velocity, the easiest thing is to decompose each velocity into the x and y coordinate system, then add each velocity
let's break down the speeds
cos 29 = v₁ₓ / v₁
sin 29 = / v₁
v₁ₓ = v₁ cos 29
v_{1y} = v₁ sin 29
v₁ₓ = 63 cos 29 = 55.10 m / s
v_{1y} = 63 sin 29 = 30.54 m / s
speed 2
cos 22 = v₂ₓ / v₂
sin 22 = v_{2y} / v₂
v₂ₓ = v₂ cos 233
v_{2y} = v₂ sin 233
v₂ₓ = 22 cos 233 = -13.24 m / s
v_{2y} = 22 sin 233 = -17.57 m / s
speed 3
cos 321 = v₃ₓ / v₃
sin 321 = v_{3y} / v₃
v₃ₓ = v₃ cos 321
v_{1y} = vₐ sin 321
v₃ₓ = 98 cos 321 = 76.16 m / s
v_{3y} = 98 sin 321 = -61.67 m / s
We already have all the component of the speeds, the resulting speed is
vₓ = v₁ₓ + v₂ₓ + v₃ₓ
vₓ = 55.10 -13.24 +76.16
vₓ = 118.02 m / s
v_{y} = v_{1y} + v_{2y} + v_{3y}
v_{y} = 30.54 -17.54 - 61.67
v_{y} = -48.67 m / s
there are two ways to give the result
v = (118.02 i -48.67 j) m / s
or in the form of magnitud and angle.
We use the Pythagorean theorem for the module
v = √ (vₓ² + v_{y}²)
v = RA (118.02² + 48.67²)
v = 127.66 m / s
let's use trigonometry for the angle
tan θ = v_{y} / vₓ
θ = tan⁻¹ (v_{1} / vₓ)
θ = tan⁻¹ (-48.67 / 118.02)
θ = -22.41
if we want to measure the angles with respect to the positive side of the x axis
θ’= 360 - 22.41
θ’= 337.59
