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4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.
5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.
If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.
<h2>
Its velocity when it crosses the finish line is 117.65 m/s</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = ?
Time, t = 6.8 s
Displacement, s = 1/4 mi = 400 meters
Substituting
s = ut + 0.5 at²
400 = 0 x 6.8 + 0.5 x a x 6.8²
a = 17.30 m/s²
Now we have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = ?
Time, t = 6.8 s
Acceleration, a = 17.30 m/s²
Substituting
v = u + at
v = 0 + 17.30 x 6.8
v = 117.65 m/s
Its velocity when it crosses the finish line is 117.65 m/s
Explanation:
It is given that, the position of a particle as as function of time t is given by :
Let v is the velocity of the particle. Velocity of an object is given by :
![v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%5B%288t%2B9%29i%2B%282t%5E2-8%29j%2B6tk%5D%7D%7Bdt%7D)
So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :
![a=\dfrac{d[8i+4tj+6k]}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5B8i%2B4tj%2B6k%5D%7D%7Bdt%7D)
At t = 0, 
Hence, this is the required solution.
Answer:
0.108 rad/s².
Explanation:
Given that
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 0.5 rev/s
We know that
1 rev/s = 6.28 rad/s
ωf= 3.14 rad/s
t= 28.9 s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
3.14 = 0 + α x 28.9
Therefore the acceleration will be 0.108 rad/s².
Therefore the answer will be 0.108 rad/s².
