Two planets follow a circular orbit around a central star in the same plane. The distance between the star at point S and one planet at point A is 135 million miles. The distance between the star and the other planet at point B is 100 million miles. If an observer on the first planet at point A sights the second planet such that ∠ S A B = 42 ° find the distance between the planets. Round to the nearest million miles.
Two planets follow a circular orbit around a central star in the same plane. The distance between the star at point S and one planet at point A is 135 million miles. The distance between the star and the other planet at point B is 100 million miles. If an observer on the first planet at point A sights the second planet such that ∠ S A B = 42 ° find the distance between the planets. Round to the nearest million miles.
Solution Summary: The author calculates the distance between both the planets by drawing a rough sketch for the conditions and labeling the sides and angles.
Two planets follow a circular orbit around a central star in the same plane. The distance between the star at point
S
and one planet at point
A
is
135
million miles. The distance between the star and the other planet at point
B
is
100
million miles. If an observer on the first planet at point
A
sights the second planet such that
∠
S
A
B
=
42
°
find the distance between the planets. Round to the nearest million miles.
The bracket BCD is hinged at C and attached to a control cable at B. Let F₁ = 275 N and F2 = 275 N.
F1
B
a=0.18 m
C
A
0.4 m
-0.4 m-
0.24 m
Determine the reaction at C.
The reaction at C
N Z
F2
D
The correct answer is C,i know that we need to use stokes theorem and parametrize the equations then write the equation F with respect to the curve but i cant seem to find a way to do it, the integral should be from 0 to 2pi but i might be wrongcould you show me the steps to get to 18pi
A 10-ft boom is acted upon by the 810-lb force as shown in the figure.
D
6 ft
6 ft
E
B
7 ft
C
6 ft
4 ft
W
Determine the tension in each cable and the reaction at the ball-and-socket joint at A.
The tension in cable BD is
lb.
The tension in cable BE is
lb.
The reaction at A is (
lb) i +
Ib) j. (Include a minus sign if necessary.)
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