Answer:
Explanation:
<u>Given Data:</u>
Radius = r = 40 m
Speed = v = 25 m/s
<u>Required:</u>
Centripetal Acceleration = 
= ?
<u>Formula:</u>
<u>Solution:</u>
![\sf a_{c} = \frac{v^2}{r} \\\\a_{c} = \frac{(25)^2}{40} \\\\a_{c} = \frac{625}{40} \\\\a_{c} = 15.6 m/s^2\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%20a_%7Bc%7D%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20%5C%5C%5C%5Ca_%7Bc%7D%20%3D%20%5Cfrac%7B%2825%29%5E2%7D%7B40%7D%20%5C%5C%5C%5Ca_%7Bc%7D%20%3D%20%5Cfrac%7B625%7D%7B40%7D%20%5C%5C%5C%5Ca_%7Bc%7D%20%3D%2015.6%20m%2Fs%5E2%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Answer: the angular frequency is 2.31 rad/s
Explanation:
The data we have is:
Radial acceleration A = 27.9 m/s^2
Beam length r = 5.21m
The radial acceleration is equal to the velocity square divided the radius of the circle (the lenght of the beam in this case)
And we can write the velocity as:
v = w*r where r is the radius of the circle, and w is the angular frequency.
w = 2pi*f
where f is the "normal" frequency.
So we have:
A = (v^2)/r = (r*w)^2/r = r*w^2
We can replace the values and find w.
27.9m/s^2 = 5.21m*w^2
√(27.9/5.21) = w = 2.31 rad/s
Making a wire thicker has the same effect as making a road wider. It makes it easier for the electron traffic to flow. The resistance decreases, and the current (traffic) increases.
Answer:
668 bright fringes
Explanation:
t = Thickness = 0.2 mm
= Wavelength = 600 nm
m = Number of fringes
We have the fringe width relation
So, total number of fringes will be including m = 0 is 
