Answer:
35mc
Step-by-step explanation:
The answer is Hx = ½ Wsin θ cos θ
The explanation for this is:
Analyzing the torques on the bar, with the hinge at the axis of rotation, the formula would be: ∑T = LT – (L/2 sin θ) W = 0
So, T = 1/2 W sin θ. Analyzing the force on the bar, we have: ∑fx = Hx – T cos θ = 0Then put T into the equation, we get:∑T = LT – (L/2 sin θ) W = 0
Let AB be a chord of the given circle with centre and radius 13 cm.
Then, OA = 13 cm and ab = 10 cm
From O, draw OL⊥ AB
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = ½AB = (½ × 10)cm = 5 cm
From the right △OLA, we have
OA² = OL² + AL²
==> OL² = OA² – AL²
==> [(13)² – (5)²] cm² = 144cm²
==> OL = √144cm = 12 cm
Hence, the distance of the chord from the centre is 12 cm.
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Ahh, basic shapes. Split up the weird shapes into easier ones. #1 can be truned into 2 trapezoids. #2 can be turned into 2 circles. #3 is a triangle and a trapezoid. #4 is 2 right triangles. #5 is a rectangel and a traingle. Finally, #6 is 3 traigles and a rectangle. Do you see how we broke the hard shapes into easier shapes?
Answer:
8 out of 11
Step-by-step explanation:
