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
<h2>
Answer: faster </h2>
The speed of sound varies depending on the medium through which the sound waves travel. In addition, it varies with changes in the temperature of the medium. This is because an <u>increase in temperature means that the frequency of interactions between the particles that transport the vibration increases</u>, hence this increase in activity increases the speed. That is why the speed of sound in a gas is not constant, but depends on the temperature.
So, if we want <u>the speed of sound in a gas to increase</u>, the<u> temperature</u> of that gas must <u>increase</u>, as well.
For example, the higher the air temperature, the greater the velocity of propagation. Experiments have shown that the speed of sound in air increases for every increase in temperature.
Therefore:
<h2>The speed of sound will be faster than in December</h2>
Answer:
0.018 J
Explanation:
The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by
where
is the magnitude of the charge
is the potential difference between point P and infinity
Substituting into the equation, we find
Answer:
Explanation:
Wave Period and Frequency In this video Paul Andersen explains how the period is the time between wave and the frequency is the number of ... Your browser does not currently recognize any of the video formats available. ... Show less Show more ... Waves: Light, Sound, and the nature of Reality.