Answer:
34.6 m/s
Explanation:
From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg
Final mass will be 31.5+25.9=57.4 kg
From formula of momentum
M1v1=m2v2
Making v2 the subject of the formula then
![V2=\frac {M1v1}{m2}](https://tex.z-dn.net/?f=V2%3D%5Cfrac%20%7BM1v1%7D%7Bm2%7D)
Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then
![V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s](https://tex.z-dn.net/?f=V2%3D%5Cfrac%20%7B100.2%20kg%5Ctimes%2019.8%20m%2Fs%7D%7B57.4%20kg%7D%3D34.56376%20m%2Fs%5Capprox%2034.6%20m%2Fs)
Answer:
the outer covering of a bird is usually for camoflauge (idk how to spell) or to attract mates and the outer part of a tree is used for protection i think, correct me if im wrong ;-;
Explanation:
Answer:
I believe the answer is It increases
Answer:
3 km/h
Explanation:
Let's call the rowing speed in still water x, in km/h.
Rowing speed in upstream is: x - 2 km/h
Rowing speed in downstream is: x + 2 km/h
It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:
8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)
Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)
8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)
Dividing by 8
(x+2) + (x-2) = (6/5)*(x² - 4)
2*x = (6/5)*x² - 24/5
0 = (6/5)*x² - 2*x - 24/5
Using quadratic formula
![x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B%28-2%29%5E2%20-%204%286%2F5%29%28-24%2F5%29%7D%7D%7B2%286%2F5%29%7D)
![x = \frac{2 \pm 5.2}{2.4}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B2%20%5Cpm%205.2%7D%7B2.4%7D)
![x_1 = \frac{2 + 5.2}{2.4}](https://tex.z-dn.net/?f=x_1%20%3D%20%5Cfrac%7B2%20%2B%205.2%7D%7B2.4%7D)
![x_1 = 3](https://tex.z-dn.net/?f=x_1%20%3D%203)
![x_2 = \frac{2 - 5.2}{2.4}](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B2%20-%205.2%7D%7B2.4%7D)
![x_2 = -1\; 1/3](https://tex.z-dn.net/?f=x_2%20%3D%20-1%5C%3B%201%2F3)
A negative result has no sense, therefore the rowing speed in still water was 3 km/h
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