Saphina AlMatary

A ball of mass 1.5kg is struck by a bat in the opposite direction to the motion of the ball. Before the impulse the ball is travelling at 16ms^-1 and the impulse of the bat on the ball is 50 Ns. Find the velocity of the ball immediately after the impact.

A vehicle of mass 600kg, moving at a speed of 30ms^-1 collides with a stationary vehicle of mass 900kg. The two vehicles lock together on impact. Calculate: 1) The speed of the two vehicles immediately after the impact. 2) The force of the impact, if the...

An object of mass 12kg is pushed 34m up a frictionless slope that is at an angle of 56^o to the horizon. Determine both (a) the work on the object and (b) its change on gravitational potential energy.

1a) Solve each of these questions by factorising the left hand side into a product of linear factors: (i) x^2 - 6x - 40 = 0 (ii) 3x^2 + 20x - 7 = 0 For each question in part (a), check your solutions by substituting them back in to the equation.

Rewrite each of the following expressions as a difference of two squares by completing the square (e.g. x^2 + 4x is rewritten (x + 2)^2 - 4: (a) x^2 - 6x (b) 3x^2 + 20x

Solve the simultaneous questions: (a) x+y+z=-3 (1)(b)x+2y+3z=-14 (1) x-y-2z=13(2) 2x+3y-z=13 (2) y-4z=25 (3) 3x-y+2z=-7 (3)

A ball of mass 2kg travelling in a straight line at 4ms^-1 is acted on by a force of 3N acting in the direction of motion for 5 seconds. What is the speed of the ball after 5 seconds? vf=vi+at but a=force/mass a = 3÷2= 1.5 vf=4+1.5×5= vf=4+7.5 vf=11.5m/s^...

Solve the simultaneous questions: (a) x+y+z=-3 (1) x-y-2z=13 (2) y-4z=25 (3) (b)x+2y+3z=-14 (1) 2x+3y-z=13 (2) 3x-y+2z=-7 (3)

Determine the nature of the roots of the following quadratic equations: (a) 3x^2 - 6x + 7 = 0 (b) 3x^2 + 6x - 7 = 0 (c) -5x^2- x/5 = 0 (d) 4x^2 - 28 49 ----x + ----- = 0 3 9

for a I got: 30.0×10.0+0.5×(3.00+10.0)^2=384.5 so the final speed is 384.5 m s??? From b to d its extremely difficult to make an equation for! Please help me asap, Kind Regards

I don't understand any of this because I don't get how you would substitute the values into the equation?

I would assume g as being 10ms^-2 therefore for the maximum height, I got: h = 1/2g (v sinθ)^2 =0.5×10^-2(30sin(30))^2 =1.125 Is this correct? How would I do the other two please?