Answer:
Explanation:
given,
diameter of merry - go - round = 2.40 m
moment of inertia = I = 356 kg∙m²
speed of the merry- go-round = 1.80 rad/s
mass of child = 25 kg
initial angular momentum of the system
final angular momentum of the system
from conservation of angular momentum
20cm spanner is suitable to open this nut , Because we feel easy to apply force at the end of spanner to rusted nut. It means that as the perpendicular distance between force and axis of rotation increase the magnitude of moment also increase.
I think the Answer will helps you.....
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Answer:
a) v_{p} = 2.83 m / s , b) 50.5º north east
Explanation:
This is a vector problem.
The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground
To make the sum we decompose the speed of the ball in its components
The angle of 30 east of the south, measured from the positive side of the x axis is
θ = 30 + 270 = 300
=
cos 300
= v_{b} sin. 300
v_{bx} = 3.60 cos 300 = 1.8 m / s
v_{by} = 3.60 sin 300 = -3,118 m / s
Let's add speeds on each axis
X axis
vₓ = v_{bx}
vₓ = 1.8 m / s
Y Axis
= v1 - vpy
v_{y} = 5.30 - 3.118
v_{y} = 2.182 m / s
The magnitude of the velocity can be found using the Pythagorean theorem
= √ (vₓ² + v_{y}²)
v_{p} = √ (1.8² + 2.182²)
v_{p} = 2,829 m / s
v_{p} = 2.83 m / s
b) for direction use trigonometry
tan θ = / vₓ
θ = tan ⁺¹ v_{y} / vₓ
θ = tan⁻¹ 2.182 / 1.8
Tea = 50.48º
This address is 50.5º north east