At the instant a 2.0-kg particle has a velocity of 4.0 m/s in the positive x direction, a 3.0-kg particle has a velocity of 5.0
1 answer:
Answer:
the speed of the center of mass of the two-particle system is 3.4 m/s.
Explanation:
Given that,
Mass of particle 1, m = 2 kg
Velocity of the particle 1, v = 4 m/s in +x direction
Mass of particle 2, m' = 3 kg
Velocity of the particle 2, v' = 5 m/s in +y direction.
The x -coordinate of velocity of the centre of mass is given by :
The y -coordinate of velocity of the centre of mass is given by :
So, the the speed of the center of mass of the two-particle system given by :
So, the speed of the center of mass of the two-particle system is 3.4 m/s.
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The tension in the swing's chain at the bottom of the swing is 178.35 N.
The given parameters:
- Mass of the girl, m = 27 kg
- Speed of the girl, v = 3 m/s
- Radius of the circle, r = 4 m
The tension in the swing's chain at the bottom of the swing is calculated as follows;
Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.
Learn more about tension in vertical circle here: brainly.com/question/19904705
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
