Answer:
the distance in meters traveled by a point outside the rim is 157.1 m
Explanation:
Given;
radius of the disk, r = 50 cm = 0.5 m
angular speed of the disk, ω = 100 rpm
time of motion, t = 30 s
The distance in meters traveled by a point outside the rim is calculated as follows;
![\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega%20t%5C%5C%5C%5C%5Ctheta%20%3D%20%28100%20%5Cfrac%7Brev%7D%7B%5Cmin%7D%20%20%5Ctimes%20%5Cfrac%7B2%5Cpi%20%5C%20rad%7D%7B1%20%5C%20rev%7D%20%5Ctimes%20%5Cfrac%7B1%5Cmin%7D%7B60%20s%7D%20%29%20%5Ctimes%20%2830%20s%29%5C%5C%5C%5C%5Ctheta%20%3D%20100%20%5Cpi%20%5C%20rad%5C%5C%5C%5Cd%20%3D%20%5Ctheta%20r%5C%5C%5C%5Cd%20%3D%20100%5Cpi%20%20%5C%20%5Ctimes%20%5C%200.5m%5C%5C%5C%5Cd%20%3D%2050%20%5Cpi%20%5C%20m%20%3D%20157.1%20%5C%20m)
Therefore, the distance in meters traveled by a point outside the rim is 157.1 m
Density<span> is the </span>mass<span> of an object </span>divided<span> by its </span>volume<span>. So the answer would be Yes. Hope it helps! (:</span>
B) not work ,because the water would freeze
The hull type that is best for use on ponds, small lakes and calm rivers is Flat Bottom Hull.
A flat bottomed boat is a boat with a flat bottomed, two-chined hull, which allows it to be used in shallow bodies of water, such as rivers, because it is less likely to ground. The flat hull also makes the boat more stable in calm water.
Answer:
The total displacement is 102 km
north of east.
Explanation:
We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.
![d_n=34km+79km*sin(35^o)=79km](https://tex.z-dn.net/?f=d_n%3D34km%2B79km%2Asin%2835%5Eo%29%3D79km)
and
![d_e=79*cos(35^o)=65km](https://tex.z-dn.net/?f=d_e%3D79%2Acos%2835%5Eo%29%3D65km)
The total displacement is given by:
![D=\sqrt{(79km)^2+(65km)^2}=102km](https://tex.z-dn.net/?f=D%3D%5Csqrt%7B%2879km%29%5E2%2B%2865km%29%5E2%7D%3D102km)
with an agle of:
![\alpha =arctg(\frac{79km}{65km})=51^o](https://tex.z-dn.net/?f=%5Calpha%20%3Darctg%28%5Cfrac%7B79km%7D%7B65km%7D%29%3D51%5Eo)