i think the answer is D.
heating surface water causes it to evaporate and enter the air as a gas.
Answer:
The speed of Susan is 2.37 m/s
Explanation:
To visualize better this problem, we need to draw a free body diagram.
the work is defined as:
![W=F*d*cos(\theta)](https://tex.z-dn.net/?f=W%3DF%2Ad%2Acos%28%5Ctheta%29)
here we have the work done by Paul and the friction force, so:
![W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J](https://tex.z-dn.net/?f=W_p%3DF_p%2Ad%2Acos%280%29%5C%5CF_p%3D30N%2Acos%2830%5Eo%29%3D26N%5C%5CW_p%3D26%2A3%2A%281%29%3D78J)
![W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J](https://tex.z-dn.net/?f=W_f%3DF_f%2Ad%2Acos%28180%29%5C%5CF_f%3D%5C%C2%B5%2A%2810%2A9.8-30N%2Asin%2830%5Eo%29%29%3D16.6N%5C%5CW_p%3D16.6%2A3%2A%28-1%29%3D50J)
Now the change of energy is:
![W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s](https://tex.z-dn.net/?f=W_p-W_f%3D%5Cfrac%7B1%7D%7B2%7Dm%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2%2878J-50J%29%7D%7B10kg%7D%7D%5C%5Cv%3D2.37m%2Fs)
Answer:
96%
Explanation
Let A the total area of the galaxy, is modeled as a disc:
A = πR^2 = π (25 kpc)^2
And let a be the area that astronomers are able to see:
a = πr^2 = π(5 kpc)^2
The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:
P = 100 a/A = (5/25)^2 = 100/25 = 4%
Therefore, the percentage of the galaxy not included, i.e. not seen is:
(100-4)% = 96%
<span>ultraviolet
Have a great day!</span>
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h