Answer:
a) d = 30.79 m
, b) θ = -22.4°
, θ = 22.4 South of East
Explanation:
The easiest way to solve problems with vectors is to use their components, for this the East-West direction coincides with the x-axis and the North-South direction coincides with the y-axis
Let's use the index for / Ricardo and the index for Jane, let's break down the displacements
Richard
X axis
x₁ = 26.0 sin (60)
x₁ = -22.52 m
Y Axis
y₁ = 26.0 cos 60
y₁ = 13 m / s
Jane
X axis
x₂ = 16.0 cos (180 +30)
x₂ = -13.85 m
Y Axis
y₂ = 16.0 sin (180 + 30)
y₂ = - 8.0 m
Now we can use Pythagoras' theorem to find the distance between them
d = √ [(x₂ -x₁)² + (y₂ -y₁)²]
d = √ [(-13.85 + 22.52)² + (-8 -13)²]
d = 30.79 m
Let's use trigonometry to enter the address
tan θ = Δy / Δx
θ = tan⁻¹ Δy / Δx
θ = tan⁻¹ (-13.85 + 22.52) / (-8 - 13)
θ = tan⁻¹ (-8.67 / 21)
θ = -22.4°
The negative sign indicates that the angle is measured from the axis clockwise.
In the form of cardinal s point is
θ = 22.4 South of East
As the time of motion increases, the velocity of the object increases downwards.
The given parameters;
- <em>acceleration due to gravity, g = -9.81 m/s²</em>
- <em>velocity of the object, v₀ = 9.8 m/s</em>
The final velocity of the object at different time is calculated as follows;
<em>when the time = 1 second;</em>
v = v₀ - gt
v = 9.8 - 9.8(1)
v = 0
<em>when the time = 2 second;</em>
v = v₀ - gt
v = 9.8 - 9.8(2)
v = -9.8 m/s
<em>when the time = 3 second;</em>
v = v₀ - gt
v = 9.8 - 9.8(3)
v = -19.6 m/s
Thus, we can conclude that as the time of motion increases, the velocity of the object increases downwards.
Learn more here:brainly.com/question/16782759
Answer:
Option C. ⁰₊₁e
Explanation:
From the question given above, the following data were obtained:
¹⁵₈O —> ¹⁵₇N + __
Let the unknown be ʸₓM
Thus, the equation becomes
¹⁵₈O —> ¹⁵₇N + ʸₓM
Next, we shall determine x, y and M. This can be obtained as follow:
8 = 7 + x
Collect like terms
8 – 7 = x
1 = x
x = 1
15 = 15 + y
Collect like terms
15 – 15 = y
0 = y
y = 0
ʸₓM => ⁰₁M => ⁰₊₁e
Therefore,
¹⁵₈O —> ¹⁵₇N + ʸₓM
¹⁵₈O —> ¹⁵₇N + ⁰₊₁e
The average velocity of the beetle is equal to the ratio between the displacement and the time:
The displacement of the beetle is
The time is t=12.8 s
Therefore, the average velocity is
in the positive x-direction.
