"The movement of water into a nutrient-rich region of the phloem decreases the pressure in that region" is the statement that is not true according <span>to the pressure-flow hypothesis. The correct option among all the options that are given in the question is the fourth option or the last option. I hope it helps you.</span>
Answer:
is reflected back into the region of higher index
Explanation:
Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.
According to Snell's law, refraction of ligth is described by the equation
where
n1 is the refractive index of the first medium
n2 is the refractive index of the second medium
is the angle of incidence (in the first medium)
is the angle of refraction (in the second medium)
Let's now consider a situation in which
so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as
Where 
is a number greater than 1. This means that above a certain value of the angle of incidence , the term on the right can become greater than 1. So this would mean
But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by
Answer:
Here, 5% of energy wasted as a heat lost or work done against frictional forces.
Explanation:
No engine has yet made that can convert its all input energy into mechanical energy. Some of the input energy goes in vain in the form of heat and friction.
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as
C = B × log₂(1 + S/N)
where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.
Since the given SN ratio is in decibels, we must first express it as a ratio with no units as
SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000
Now that we have S/N, we can solve for its capacity (in bits per second) as
C = 4000 × log₂(1 + 1000)
C = 39868.91 bps
Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.
Answer: 40 kbps
