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A ball is dropped from rest at a height of 2 metres. Assuming acceleration due to gravity (g) is 10m/s^2 calculate the velocity of the ball just before it hits the floor.

Using the equation V² = U² + 2as, where V = final velocity, U = initial velocity, a = acceleration, s = displacement, we can substitute in the values given in the question t...

Answered by Phil B. • Physics tutor

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Difference between Qui/Que?

Qui and Que are relative pronouns. That means that they are used to form one sentence out of two different ones to avoid repetition. Although both pronouns serve the same purpose, they a...

Answered by Eoghan L. • French tutor

1769 Views

When a particle travels in a circle of radius r, at constant speed v, what is its acceleration

v²/r, towards the center of the circle.Remember that acceleration is the rate of change of velocity, not merely of speed. This means that the change in direction is im...

Answered by • Physics tutor

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Find the stationary points on the curve: y = x^3 + 3x^2 +2x+5

Firstly differentiate the function:f(x) = x³ + 3x² + 2x + 5 (function)f'(x) = 3x² + 6x + 2 (gradient function)
Stationary points are points where the graph ...

Answered by Nicolas C. • Maths tutor

5366 Views

Work out the gradient of the curve y=x^3(x-3) at the point (3,17)

First simplify the equation of the curve y= x^4 - 3x^3 .The gradient is the differential.To differentiate, bring down the power and take one from it.x^4 becomes 4x^3-3x^3 becomes (-3x3)= -9x^2dy/dx= 4x^3 ...

Answered by Sophie M. • Further Mathematics tutor

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