Λ = 3*10^8 / 9*10^8 = 1/3 m
no. of wavelengths = 60/(1/3) = 180
Answer:
total length of the spiral is L is 5378.01 m
Explanation:
Given data:
Inner radius R1=2.5 cm
and outer radius R2= 5.8 cm.
the width of spiral winding is (d) =1.6 \mu m = 1.6x 10^{-6} m
the total length of the spiral is L is given as



= 5378.01 m
Answer:
The rate of transfer of energy is equal to 23.76W or 23.76J/s as may be required both forms are correct. The physical quantities needed to calculate the rate of energy transfer are the linear mass density or mass per unit length, tension force, amplitude, angular frequency( which is equal to 2pi •f )
Explanation:
The required quantity is the average power or average rate of energy transfer which differs from the maximum or instantaneous rate of energy transfer. The calculation steps to the answer above can be found in the attachment below. Should the requested quantity be the instantaneous quantity the answer will be 2 x Pav which equals 47.52W or 47.52J/s.
Answer:
The speed in the first point is: 4.98m/s
The acceleration is: 1.67m/s^2
The prior distance from the first point is: 7.42m
Explanation:
For part a and b:
We have a system with two equations and two variables.
We have these data:
X = distance = 60m
t = time = 6.0s
Sf = Final speed = 15m/s
And We need to find:
So = Inicial speed
a = aceleration
We are going to use these equation:


We are going to put our data:


With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.



![[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}](https://tex.z-dn.net/?f=%5B%5Csqrt%7B%2815m%2Fs%29%5E2-%282%2Aa%2A60m%29%7D%5D%5E%7B2%7D%3D%5B15m%2Fs-%28a%2A6s%29%5D%5E%7B2%7D)








If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2
After, we are going to determine the speed in the first point:




For part c:
We are going to use:




Answer:
Momentum, p = 7.2 g-m/s
Explanation:
It is given that,
The momentum of an object is 
We need to express momentum in any equivalent units. There can be many solutions of this problem. Some of the units of mass are gram, milligram etc. units of length are meters, mm etc.
Since, 1 kg = 1000 gram
So, 
Therefore, the momentum of the object is 7.2 g-m/s. Hence, this is the required solution.