Answer:
a lot my dude
Explanation:
cars are heavy
Write an equation that expresses the following relationship. u varies jointly with p and d and inversely with w In your equation, use k as the constant of proportionality.
[tex]\boxed{y=k\frac{pd}{w}}[/tex]
Explanation:Let's explain what direct and indirect variation mean:
When we say that [tex]y[/tex] varies jointly as [tex]x \ and \ w[/tex], we mean that:[tex]y=kxw[/tex] for some nonzero constant [tex]k[/tex] that is the constant of variation or the constant of proportionality.
On the other hand, when we say that [tex]y[/tex] varies inversely as [tex]x[/tex] or [tex]y[/tex] is inversely proportional to [tex]x[/tex], we mean that:[tex]y=\frac{k}{x}[/tex] for some nonzero constant [tex]k[/tex], where [tex]k[/tex] is also the constant of variation.
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In this problem, [tex]u[/tex] varies jointly with [tex]p[/tex] and [tex]d[/tex] and inversely with [tex]w[/tex], being [tex]k[/tex] the constant of proportionality, then:
[tex]\boxed{y=k\frac{pd}{w}}[/tex]
The equation expressing the described relationship is u = kpd/w. This represented u varying jointly with p and d and inversely with w, with k being the constant of proportionality.
Explanation:The equation that expresses the relationship where 'u' varies jointly with 'p' and 'd' and inversely with 'w' using 'k' as the constant of proportionality would be: u = kpd/w. Here, 'k' is the constant of proportionality which determines the rate at which 'u' varies relative to 'p', 'd', and 'w'. Joint variability is expressed by multiplying the variables 'p' and 'd', while inverse proportionality is shown through dividing by 'w'.
This equation effectively expresses the described relationship - as 'p' or 'd' increase (or 'w' decrease), 'u' increases and vice versa. It’s important to understand that the constant 'k' would need to be determined based on additional context or data.
Learn more about Joint and Inverse Variations here:https://brainly.com/question/29181669
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A Ferris wheel has diameter of 10 m and makes one revolution in 8.0 seconds. A woman weighing 670 N is sitting on one of the benches attached at the rim of the wheel. What is the net force on this woman as she passes through the highest point of her motion?
Answer:
208 N
Explanation:
The net force on the woman is equal to the centripetal force, which is given by
[tex]F=m\frac{v^2}{r}[/tex]
where
m is the mass of the woman
v is her speed
r is the radius of the wheel
here we have:
r = d/2 = 5 m is the radius of the wheel
[tex]m=\frac{W}{g}=\frac{670 N}{9.8 m/s^2}=68.4 kg[/tex] is the mass of the woman (equal to her weight divided by the acceleration of gravity)
The wheel makes one revolution in t=8.0 s, so the speed is:
[tex]v=\frac{2\pi r}{t}=\frac{2\pi (5.0 m)}{8.0 s}=3.9 m/s[/tex]
so now we can find the centripetal force:
[tex]F=(68.4 kg)\frac{(3.9 m/s)^2}{5.0 m}=208 N[/tex]
A resultant vector is 5 units long and makes an angle of 23 degrees measured counter-clockwise with respect to the positive x-axis. What are the magnitude and angle (measured counter-clockwise with respect to the positive x-axis) of the equilibrant vector?
Answer:
5 units at 203 degrees
Explanation:
The equilibrant vector must have components that are opposite to those of the initial vector.
The components of the initial vector are:
[tex]v_x = v cos \theta = 5 cos 23^{\circ}=4.60\\v_y = v sin \theta = 5 sin 23^{\circ} = 1.95[/tex]
So the components of the equilibrant vector must be
[tex]v_x = -4.60\\v_y = -1.95[/tex]
which means its magnitude is
[tex]v=\sqrt{v_x^2 + v_y^2}= 5[/tex] (same magnitude as the initial vector)
and it is located in the 3rd quadrant, so its angle will be
[tex]\theta = 180^{\circ} + tan^{-1} (\frac{1.95}{4.60})=203^{\circ}[/tex]