Answer:
I think it's 4 but don't hold me to it
Step-by-step explanation:
Part I)
The module of vector AB is given by:
lABl = root ((- 3) ^ 2 + (4) ^ 2)
lABl = root (9 + 16)
lABl = root (25)
lABl = 5
Part (ii)
The module of the EF vector is given by:
lEFl = root ((5) ^ 2 + (e) ^ 2)
We have to:
lEFl = 3lABl
Thus:
root ((5) ^ 2 + (e) ^ 2) = 3 * (5)
root ((5) ^ 2 + (e) ^ 2) = 15
Clearing e have:
(5) ^ 2 + (e) ^ 2 = 15 ^ 2
(e) ^ 2 = 15 ^ 2 - 5 ^ 2
e = root (200)
e = root (2 * 100)
e = 10 * root (2)
The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is T^2 = A^3.
Let the orbital period of planet X be T(X) and that of planet Y = T(Y) and let the mean distance of planet X from the sun be A(X) and that of planet Y = A(Y), then
A(Y) = 2A(X)
[T(Y)]^2 = [A(Y)]^3 = [2A(X)]^3
But [T(X)]^2 = [A(X)]^3
Thus [T(Y)]^2 = 2^3[T(X)]^2
[T(Y)]^2 / [T(X)]^2 = 2^3
T(Y) / T(X) = 2^3/2
Therefore, the orbital period increased by a factor of 2^3/2
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Be more specific. Like what is it asking you to do ? I'm trying to help. :)
Answer:
17
Step-by-step explanation:
7h+3
7(2)+3
14+3
=17
