To find the scientific notation, you need to divide at the decimal by the power of 10. So since there are 2 powers of 10, what you want to do is move the decimal 2 places to the left which will give you: .054
A. 9 J
In a force-distance graph, the work done is equal to the area under the curve in the graph.
In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:
where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:
And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:
B. 24.5 m/s
The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:
where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:
:<span> </span><span>The gradient of the curve 1/x at x=2 is m = -¼
We may choose any length of line to represent the direction of the slope (direction vector) at that point. We could choose a line for which x = 2 and then y would have to be -½ so that the gradient is still = -½/2 = -¼. It is simply convenient to choose a unit length for x, making y = -¼ The length of the resultant of x and y is √(1²+¼²) = √(17/16) = √(17)/4 which is a direction vector. If we had taken the direction vector to be (2, ½) then we would have a resultant direction vector of √17/2. It doesn't really matter what length the direction vector is - it's job is only to show the direction. So their choice of 1 is quite arbitrary but convenient, since it is easy to work with units – that's why we use units!
Now, we know that the magnitude of the velocity vector must be 5 and the magnitude of our direction vector at the moment is √(17)/4. We therefore need to multiply this direction vector by 20/√(17) to get 5 – just try it : √(17)/4 × 20/√(17) = 5.
We could equally well have done this with (2, ½) and would have got 2½ for lambda.</span>
