You don't need to do any calculations, you are already given that n=1.5
n is index of refraction. it's 1.5
The linear speed of the ladybug is 4.1 m/s
Explanation:
First of all, we need to find the angular speed of the lady bug. This is given by:
![\omega=\frac{2\pi}{T}](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7BT%7D)
where
T is the period of revolution
The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:
T = 1 s
Therefore, the angular speed is
![\omega=\frac{2\pi}{1 s}=6.28 rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7B1%20s%7D%3D6.28%20rad%2Fs)
Now we can find the linear speed of the ladybug, which is given by
![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
where:
is the angular speed
r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation
Substituting, we find
![v=(6.28)(0.65)=4.1 m/s](https://tex.z-dn.net/?f=v%3D%286.28%29%280.65%29%3D4.1%20m%2Fs)
Learn more about angular speed:
brainly.com/question/9575487
brainly.com/question/9329700
brainly.com/question/2506028
#LearnwithBrainly
Answer:
![\boxed{\sf Focal \ length = 10 \ cm}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%5Csf%20Focal%20%5C%20length%20%3D%2010%20%5C%20cm%7D%20)
Given:
Radius of curvature (R) of a spherical mirror = 20
To Find:
Focal length (f)
Explanation:
Formula:
![\boxed{ \bold{\sf Focal \ length \ (f) = \frac{Radius \ of \ curvature \ (R)}{2}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%5Csf%20Focal%20%5C%20length%20%5C%20%28f%29%20%3D%20%5Cfrac%7BRadius%20%5C%20of%20%5C%20curvature%20%5C%20%28R%29%7D%7B2%7D%7D%7D)
Substituting value of R in the equation:
![\sf \implies f = \frac{20}{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20f%20%3D%20%20%5Cfrac%7B20%7D%7B2%7D%20)
![\sf \implies f = \frac{ \cancel{2} \times 10}{ \cancel{2}}](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20f%20%3D%20%5Cfrac%7B%20%5Ccancel%7B2%7D%20%5Ctimes%2010%7D%7B%20%5Ccancel%7B2%7D%7D%20)
![\sf \implies f = 10 \: cm](https://tex.z-dn.net/?f=%20%5Csf%20%5Cimplies%20f%20%3D%2010%20%5C%3A%20cm)
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
![V^2 = U^2 + 2dS](https://tex.z-dn.net/?f=V%5E2%20%3D%20U%5E2%20%2B%202dS)
Substituting the values into the formula, we have;
![V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}](https://tex.z-dn.net/?f=V%5E2%20%3D%2022%20%2B%202%281.4%29%28110%29%5C%5C%5C%5CV%5E2%20%3D%2022%20%2B%20308%5C%5C%5C%5CV%5E2%20%3D%20330%5C%5C%5C%5CV%5E2%20%3D%20%5Csqrt%7B330%7D)
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
<em></em>
Read more: brainly.com/question/8898885