This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.
Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.
I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.
Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

Let the orbital period of the earth be
and its mean distance of from the sun be
.
Also let the orbital period of the planet be
and its mean distance from the sun be
.
Equation (2) therefore implies the following;

We make the period of the planet
the subject of formula as follows;

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

Substituting equation (5) into (4), we obtain the following;

cancels out and we are left with the following;

Recall that the orbital period of the earth is about 365.25 days, hence;

A joule is one Newton of force applied over a meter.
For every meter, the brakes put 240000N of force (N=Newtons).
For 40m, multiply the Newtons by 40.
240000N*40=9600000N