Answer:
e. 27 rho.
Explanation:
The density of an object, assumed constant, is defined as the relationship between the mass and the volume.
If the object is compressed in such a way that the diameter is reduced to 1/3, this means, that the radius will be reduced in the same proportion.
Now, if the mass remains the same (the compression can't change it) , the volume (assumed to be a perfect sphere) will be reduced also:
V₀ = 4/3*π*r³
Vf= 4/3*π*(r/3)³ = 4/3*π*(r³/27) = V₀/27
As the volume is in in the denominator of the expression for density, this means that the new density will be equal to 27 times the original one:
ρ₀ = m/ V₀
ρ₁ = m/ V₁ = m/ (V₀/27) = 27* (m/V₀) = 27*ρ₀
Given that,
The radius of a circular path, r = 32 m
The time of one revolution of a rider is 0.98 s.
To find,
The speed of the rider.
Solution,
Let v is the speed of the rider. Speed is equal to total distance divided by time taken.
So, the speed of the rider is 205.16 m/s.
Answer= 12.5 km
0.5 km x km
_______=______
1 min 25 min
Cross multiply
1x=12.5
divide both sides by 1
x=12.5
You traveled 12.5 km
Answer:
7500 m
Explanation:
The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.
We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so
d = 2L
In a time of
Electromagnetic waves travel in a vacuum at the speed of light, which is equal to
Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:
(1)
where
, where L is the distance between the radar and the object
Re-arranging eq(1) and substituting, we find L:
The average speed <em>appears to be</em> (distance) / (time) =
(length of the cable) / (time from when a pulse goes in until it comes out the other end) .
That's 1,200,000 meters/ 0.006 second = 2 x 10^8 = <em>2 hundred million m/sec</em>
That figure is about 66.7% of the speed of light in vacuum.
The reason I went through all of this detail was to point out that this is
NOT necessarily the speed of light in this glass, for two reasons.
1). The path of light through an optical fiber is not straight down the middle. In the original fibers of 20 or 30 years ago, the light bounced back and forth off the inside walls of the fiber, and zig-zagged its way along the length. In current modern fibers, it still zig-zags, but it's a more gentle, up-and-down curved path. In either case, the distance covered by the light inside the fiber is more than the straight length of the cable, and the time it takes it to come out the other end is more than its actual speed inside the glass would have meant if it could have traveled straight through the pipe.
2). This problem talks about an optical fiber that's 1,200km long. There is loss in optical fiber, and you're NOT going to get light all the way through a single piece of it that's something like 745 miles long. It takes electronic repeaters, "boosters", and regenerators every few miles to keep it going, and these devices add "latency" or time delay in the process of going through them. That delay in the electronics shows up as apparent delay through the fiber-optic cable, and it makes the speed through the glass appear to be slower than it actually is.
