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A block of mass 5 kg is being pushed over level ground by rod at 60 degrees to horizontal with force 40 N with acc. 1.5 what is the frictional force of the surface and draw a diagram with the forces acting on the block

Block has normal force, weight, friction and the force from the rod acting on itResultant force = 5*1.5=7.5horizontal component of force from rod = 207.5= 20 - FF= 12.5

Answered by Henry B. • Maths tutor

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Solve algebraically the simultaneous equations: 6x=5-2y 12.5=3x+3y

Rearrange both equations into the form ax + by = c. 3x+3y = 12.5 6x +2y = 5. 2) Make either x or y have the same coefficient in each equation. In this case we use x, to do this we multiply the first...

Answered by Laura W. • Maths tutor

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Expand and simplify (x – 9)(x + 2)

x^2 - 7x - 18

Answered by JOSEPH B. • Maths tutor

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Why do we need the constant of integration?

Consider three simple functions F(x)=2x, G(x)=2x-6, H(x)=2x + π/2. We can differentiate these functions with respect to x to get F’(x)=f(x)=2, G’(x)=g(x)=2, H’(x)=h(x)=2. Clearly the derivatives of these ...

Answered by Panoraia C. • Maths tutor

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Find the coordinates of the mid-point AB where A (-3,-3) and B (1,3)

X coordinates = (-3+1)/2 = (-2)/2 = -1Y coordinates = (-3+3)/2 = (0) /2 = 0 Coordinates of the midpoint = (x,y) = (-1,0)
If they were struggling to visualise it/to avoid simple mist...

Answered by Edward W. • Maths tutor

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