Can someone help me on boding elements? please im struggling. look at the picture and also if you answer can you have a picture
of it written out like it is. :)
1 answer:
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linear charge density of system of two line charges is given as
now as we know that electric field due to a line charge at some distance from it is given by
so here we will first find the electric field of first line charge at the position of other line charge
now as we know that
here q = charge on the line charge system at which force is required
E = electric field on that system of charge where force is required
now we can find the charge by
Now using the above formula
so force on the part of wire is F = 0.0811 N
Answer:
a) W = - 6.825 J, b) θ = 1.72 revolution
Explanation:
a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle
W = ΔK
W = K_f - K₀
W = ½ m v_f² - ½ m v₀²
W = ½ 0.325 (5.5² - 8.5²)
W = - 6.825 J
b) find us the coefficient of friction
Let's use Newton's second law
fr = μ N
y-axis (vertical) N-W = 0
fr = μ W
work is defined by
W = F d
the distance traveled in a revolution is
d₀ = 2π r
W = μ mg d₀ = -6.825
μ =
The total work as the object stops the final velocity is zero v_f = 0
W = 0 - ½ m v₀²
W = - ½ 0.325 8.5²
W = - 11.74 J
μ mg d = -11.74
we subtitle the friction coefficient value
( ) m g d = -11.74
6.825 = 11.74
d = 11.74/6.825 d₀
d = 1.7201 2π 0.400
d = 4.32 m
this is the total distance traveled, the distance and the angle are related
θ = d / r
θ = 4.32 / 0.40
θ = 10.808 rad
we reduce to revolutions
θ = 10.808 rad (1rev / 2π rad)
θ = 1.72 revolution
<h2>Answer: Earth's orbital path around the Sun</h2><h2></h2>
The <u>Ecliptic</u> refers to the orbit of the Earth around the Sun. Therefore, <u>for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.</u>
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It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about approximately. This is due to the inclination of the Earth's axis.
Hence, the correct option is Earth's orbital path around the Sun.
Well, the answer is obviously D, because it's the only one that starts with 3.134.
Moving the decimal place 9 times to the right will give us:
3,134,000,000
Moving the decimal place 9 times to the right will give us:
3,134,000,000