Area of low pressure where air masses meet and rise
2 answers:
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We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
where
is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance
In our problem, I=4.00 A and
, so the potential difference is
where
I is the current flowing through it
R is the resistance
In our problem, I=4.00 A and
Answer:
-9.46 N
Explanation:
In order to get the value of the x component of the resultant force, we need to get the value of the x component of each force.
This value will be the projection of the force vector, on the x-axis.
For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:
cos θ =
In this case, x = F₁ₓ, and r = F₁
θ, measured from the positive x axis counterclockwise, is as follows:
θ= 180º-55º = 125º
⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N
We can repeat the process for F₂, as follows:
For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:
cos θ =
In this case, x = F₂ₓ, and r = F₂
θ, measured from the positive x axis counterclockwise, is as follows:
θ= 180º + 53.3º = 233.3º
⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N
The total component of both forces along the x axis, can be found just adding both components, as follows:
Fₓ = F₁ₓ + F₂ₓ = -5.28 N + -4.18 N = -9.46 N
1) Analyze the equation
Vₓ = α + β t²
Vₓ = 3.00 + 0.100 t²
That is a quadratic equation, so the graph is a parabola.
2) Therefore, take into account that the main points to draw a parabola are:
i) vertex
ii) concavity
iii) y-intercepts
iv) x - intercepts
Also, for all graphs you need the domain and the range.
3) Find the y-intercept (t = 0)
t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00
4) Find the x-intercepts (Vₓ = 0)
Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²
⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.
5) Concavity
Since, the coefficient of t² is positive, the parabola open upwards.
6) Vertex
It is the local minimum of the equation. You can find it by the first derivative
Vₓ' = 2×0.100 t = 0 ⇒ t = 0
Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s
⇒ vertex = (0,3.00)
7) The domain is given t ∈ [0,5.00]
8) You can also build a table with several points in the domain
t =0; Vₓ = 3.00 + 0.100 (0)² = 0
t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10
t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40
t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90
t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60
t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50
9) Range: Vₓ ∈ [ 3.00, 5.50]
10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.
