Answer:
Magnitude = 4.056 m
Direction = 42.3⁰
Explanation:
The vector is resolved in terms of the vertical and horizontal components. Let's look each of these separately.
The vector 4.40 is directed East. This automatically becomes a horizontal component.
But we know that there is a vector 3.40 North West. The angle the vector makes with the horizontal is 61⁰.
Resolving the vectors should yield the horizontal and vertical components:
Horizontal components
The first component is 4.40 m
The second one is derived by resolving 3.40 to the horizontal like this 3.40 × - cos 61⁰ = -1.648 m
Adding the horizontal component gives 4.40 m + ( -1.648 m) = 2.752 m
Vertical components
Resolve 3.40 with the angle 61⁰ like this: vertical comp = 3.41 × sin 61
= 2.98 m
The magnitude is given by √[(2.98)²+ (2.752)²] = 4.056 m Ans
The direction us given by tan⁻¹ (2.98/2.752) = 42.3⁰ Ans
Answer:
The variable manipulated or controlled by the experimenter is called the independent variable.
Example:
If the flow velocity at the bottom of a tank is measured by varying the height of water in the tank, we are measuring velocity as a function of water height.
Therefore,
water height = independent variable (controlled)
velocity = dependent variable (measured in response to water height).
Mathematically,
v = f(h)
where v = response variable (dependent)
h = controlled variable (independent).
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
I believe d all of the above
