If you could live on the moon through one lunar cycle, how you would experience the phases of the moon?
1 answer:
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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
Answer:
horizontal direction force move wagon at 18.79 N
Explanation:
given data
force F = 20 N
angle = 20 degree
to find out
What part of the force moves the wagon
solution
we know here as per attach figure
boy pull a wagon at force 20 N at angle 20 degree
so there are 2 component
x in horizontal direction i.e F cos20
and y in vertical direction i.e F sin20
so we can say
horizontal direction force is move the wagon that is
horizontal direction force = F cos 20
horizontal direction force = 20× cos20
horizontal direction force = 18.79 N
so horizontal direction force move wagon at 18.79 N
