Red shift in the light coming towards us from other galaxies.
Answer:
Work done by the force of friction is 305.96 joules.
Explanation:
It is given that,
Mass of the child, m = 21 kg
Height of the slide, h = 1.5 m
Speed of the child at the bottom, v = 0.51 m/s
Let W is the work done by the force of friction on the child. It can be calculated using the work energy theorem as :
![mgh+W=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%2BW%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![W=\dfrac{1}{2}mv^2-mgh](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2-mgh)
W = -305.96 joules
So, the work done by the force of friction on the child is 305.96 joules. Hence, this is the required solution.
All you have to do is add 600 + 1,400 which = 2,000
So your answer is D) 2,000 J
I hope this helps. :)
I had the same test btw
Answer: no I don’t think so
Explanation:
Enumerating the data points from the chart over ten years:
{100, 105, 95, 102, 93, 105, 98, 99, 101, 100}
Their initial, final, and the median values are all 100. The mean value is 99.8, and the standard deviation slightly less than 3.9. All of these statistics indicate stability of the population over the observed span of ten years. There is neither a significant growth, nor a decline. Assuming stability is what corresponds to a species doing well in a habitat, a conclusion can be made that the species is doing well!
Provided the conditions of the artificial habitat won't change significantly in the next five years, the population will continue to remain close to an average of 100, with minor deviations of +/- 4 likely year by year.