Answer:
d’= (0.561 i ^ - 0.634 j ^) m
, d’= 0.847 m
, 48.5 south east
Explanation:
This is a displacement exercise, one of the easiest methods to solve it is to decompose the displacements in a coordinate system. Let's start with beetle 1
Let's use trigonometry to break down your second displacement
d₂ = 0.89 m θ = 32 north east
sin θ = / d₂
d_{2y} = d2 sin 32
d_{2y} = 0.89 sin 32
d_{2y} = 0.472 m
cos 32 = d₂ₓ / d₂
d₂ₓ = d₂ cos 32
d₂ₓ = 0.89 cos 32
d₂ₓ = 0.755 m
We found the total displacement of the beetle 1
X axis
d₁ = 0.58 i ^
Dₓ = d₁ + d₂ₓ
Dₓ = 0.58 + 0.755
Dₓ = 1,335 m
Axis y
D_{y} = d_{2y}
D_{y} = 0.472 m
Now let's analyze the second beetle
d₃ = 1.37 m θ = 35 north east
Sin (90-35) = d_{3y} / d₃
d_{3y} = d₃ sin 55
d_{3y} = 1.35 sin 55
d_{3y} = 1,106 m
cos 55 = d₃ₓ / d₃
d₃ₓ = d₃ cos 55
d₃ₓ = 1.35 cos 55
d₃ₓ = 0.774 m
They ask us what the second displacement should be to have the same location as the beetle 1
Dₓ = d₃ₓ + dx’
D_{y} = d_{3y} + dy’
dx’= Dₓ - d₃ₓ
dx’= 1.335 - 0.774
dx’= 0.561 m
dy’= D_{y} - d_{3y}
dy’= 0.472 - 1,106
dy’= -0.634 m
We can give the result in two ways
d’= (0.561 i ^ - 0.634 j ^) m
Or in the form of module and address
d’= √ (dx’² + dy’²)
d’= √ (0.561² + 0.634²)
d’= 0.847 m
tan θ = dy’/ dx’
θ = tan⁻¹ dy ’/ dx’
θ = tan⁻¹ (-0.634 / 0.561)
θ = -48.5
º
This is 48.5 south east
Answer:d
Explanation:
Given
First car is moving towards east with velocity 20 m/s
then it turns towards north then velocity is
suppose car takes t sec to change its path so average acceleration is given by
So average acceleration is towards North of west.
Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9 m/s²,
g = 9.81 m/s² : acceleration due to gravity
W= 75 N : Block weight
W= m*g
m = W/g = 75/9.8= 7.65 kg : Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax
20-Ff= m*ax
20-μk*100
= 7.65*(-0.90 )
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885
<span>It is traveling at 10m/s.</span>
It takes Earth about 24 hours to rotate once on its axis. The movement of one object around another object is called revolution. ... Earth's path as it revolves around the sun is called its orbit. Earth's orbit is a slightly elongated circle, or ellipse.