The answer to this questions would be nuclear engineer. The overall career of engineering would require physics as engineers need to know how the overall mechanics of nature affect the way their structure is built.
Hope this helped :D
Answer:
18.44km
Explanation:
The diagram explains better.
From the diagram,
A is the harbor;
B is Treasure Island;
C is Nomans land.
Using cosine rule, we can find the distance between C and B.
BC² = AB² + AC² - 2*AB*AC*cos(A)
AB = 26km
AC = 11km
A = 37°
BC² = 26² + 11² - (2*26*11*cos37)
BC² = 676 + 121 - (572*0.7986)
BC² = 797 - 456.8
BC² = 340.2
BC = 18.44 km
The distance traveled by Moe from Nomans land to Treasure Island is 18.44km
Answer:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
Answer:
a) ![v_{o}=4.98m/s](https://tex.z-dn.net/?f=v_%7Bo%7D%3D4.98m%2Fs)
b) ![a=1.67m/s^{2}](https://tex.z-dn.net/?f=a%3D1.67m%2Fs%5E%7B2%7D)
Explanation:
From the exercise we got final position, final velocity and how much time does it takes the body to cover the distance
![x=60m\\t=6s\\v=15m/s](https://tex.z-dn.net/?f=x%3D60m%5C%5Ct%3D6s%5C%5Cv%3D15m%2Fs)
From the concept of moving objects we know the following equations:
and ![x=x_{o}+v_{o}t+\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=x%3Dx_%7Bo%7D%2Bv_%7Bo%7Dt%2B%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
Now, we have two equations with two unknowns
(1)
Now, we need to replace (1) in the other equation
![x=x_{o}+(v-at)t+\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=x%3Dx_%7Bo%7D%2B%28v-at%29t%2B%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
![60=vt-at^{2}+\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=60%3Dvt-at%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
![60=vt-\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=60%3Dvt-%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
Solving for a
b) ![a=\frac{2(vt-60)}{t^{2} } =\frac{2((15)(6)-60)}{(6)^{2} }=1.67m/s^{2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B2%28vt-60%29%7D%7Bt%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B2%28%2815%29%286%29-60%29%7D%7B%286%29%5E%7B2%7D%20%7D%3D1.67m%2Fs%5E%7B2%7D)
Now, we can solve (1)
a) ![v_{o}=v-at=15-1.67(6)=4.98m/s](https://tex.z-dn.net/?f=v_%7Bo%7D%3Dv-at%3D15-1.67%286%29%3D4.98m%2Fs)