When two perpendicular forces 9.0 N (tow...

When two perpendicular forces 9.0 N (toward positive x) and 7.0 N (toward positive y) act on a body of mass 6.0 kg, what are the (a) magnitude and (b) direction of the acceleration of the body ?

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To solve the problem, we will follow these steps: ### Step 1: Identify the Forces The two forces acting on the body are: - \( F_1 = 9.0 \, \text{N} \) (toward positive x-axis) - \( F_2 = 7.0 \, \text{N} \) (toward positive y-axis) ### Step 2: Calculate the Resultant Force ...

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A body of mass 5kg is acted upon by two perpendicular force 8N and 6N, find the magnitude and direction the acceleration

`3 ms^(-2), theta=cos^(-1)(0.8) " from "8 m`

`2 ms^(-2), theta=cos^(-1)(0.6) " from "6 m`

`3 ms^(-2), theta=cos^(-1)(0.9) " from "6 m`

`5 ms^(-2), theta=cos^(-1)(0.81) " from "8 m`

The forces each of 20N act on a body at 120^(@) The magnitude and direction of resultant is

`20N,phi=60^(@)`

`20sqrt2N,phi=60^(@)`

`10sqrt2N,phi=0^(@)`

`10sqrt2N,phi=120^(@)`

A body of mass 10kg is acted upon by two perpendicular force, 6N . The resultant acceleration of the body is.

`1ms^(-2)` at an angle of `tan^(-1)(3/4)` w.r.t. `8N` force

`0.2ms^(-2)` at an angle of `tan^(-1)(3/4)` w.r.t. `8N` force

`1ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t. `8N` force

`0.2ms^(-2)` at an angle of `tan^(-1)(4/3)` w.r.t. `8N` force