Physics, asked by Anonymous, 29 days ago

In the above figure, Acceleration of bodies A, B and Care shown with directions. Values of b and c are w.r.t ground whereas a is acceleration of block A w.r.t Wedge C. Acceleration of block A w.r.t ground is

a)
 \sqrt{ {(b + c)}^{2}  +  {a}^{2} }
b)
c - (a + b) \cos(θ)
c)
 \sqrt{( {(b + c)}^{2}  {c}^{2} - 2(b + c) c\cos(θ)  }
d)
 \sqrt{ {(b + c)}^{2} +  {c}^{2} + 2(b + c)c \cos(θ)   }
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Answered by kaushal9006227191
1

Answer:

In the above figure, Acceleration of bodies A, B and Care shown with directions. Values of b and c are w.r.t ground whereas a is acceleration of block A w.r.t Wedge C. Acceleration of block A w.r.t ground is

a)

\sqrt{ {(b + c)}^{2} + {a}^{2} }(b+c)2+a2

b)

c - (a + b) \cos(θ)c−(a+b)cos(θ)

c)

\sqrt{( {(b + c)}^{2} {c}^{2} - 2(b + c) c\cos(θ) }((b+c)2c2−2(b+c)ccos(θ)

d)

\sqrt{ {(b + c)}^{2} + {c}^{2} + 2(b + c)c \cos(θ) }(b+c)2+c2+2(b+c)ccos(θ)

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Answered by Anonymous
5

Answer:

refer the attachment.

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