Answer:
C ) 1.53
Explanation:
The critical angle of a material is given by the formula
![sin c = \frac{1}{n}](https://tex.z-dn.net/?f=sin%20c%20%3D%20%5Cfrac%7B1%7D%7Bn%7D)
where
c is the critical angle
n is the refractive index
This formula is valid if the second medium is air (which is the case of the problem).
In this problem, we know the critical angle:
![c=40.8^{\circ}](https://tex.z-dn.net/?f=c%3D40.8%5E%7B%5Ccirc%7D)
Therefore we can rearrange the equation to find the refractive index:
![n=\frac{1}{sin c}=\frac{1}{sin 40.8^{\circ}}=1.53](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B1%7D%7Bsin%20c%7D%3D%5Cfrac%7B1%7D%7Bsin%2040.8%5E%7B%5Ccirc%7D%7D%3D1.53)
Answer:
The average linear velocity (inches/second) of the golf club is 136.01 inches/second
Explanation:
Given;
length of the club, L = 29 inches
rotation angle, θ = 215⁰
time of motion, t = 0.8 s
The angular speed of the club is calculated as follows;
![\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%28%5Cfrac%7B%5Ctheta%7D%7B360%7D%20%5Ctimes%202%5Cpi%2C%20%5C%20rad%29%20%5Ctimes%20%5Cfrac%7B1%7D%7Bt%7D%20%5C%5C%5C%5C%5Comega%20%3D%20%20%28%5Cfrac%7B215%7D%7B360%7D%20%5Ctimes%202%5Cpi%2C%20%5C%20rad%29%20%5Ctimes%20%5Cfrac%7B1%7D%7B0.8%20%5C%20s%7D%20%5C%5C%5C%5C%5Comega%20%3D%204.69%20%5C%20rad%2Fs)
The average linear velocity (inches/second) of the golf club is calculated as;
v = ωr
v = 4.69 rad/s x 29 inches
v = 136.01 inches/second
Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second
Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V
<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.
The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h. No rounding is necessary.</span>
Answer:
a. 0.18Hz
b. 0.56m/s
Explanation:
From the question we can deduct the following parameters
The wavelength, λ is define as the distance between two successful crest or trough and from the question we conclude that wavelength is 3.17m.
Also the period of the wave T can be computed as
T=22.6/4
T=5.65secs.
a. To compute the frequency, recall that frequency, F=1/period.
Hence,
F=1/5.65
F=0.18Hz
b. Next we compute the wave speed.
Wave speed=frequency *wavelength
Wave speed =0.18*3.17
Wave speed =0.56m/s