Answer:
Hence the pressure is 
Explanation:
Given data
Q=1500 J system gains heat
ΔV=- 0.010 m^3 there is a decrease in volume
ΔU= 4500 J internal energy decrease
We know work done is
W= Q- ΔU
=1500-4500= -3000 J
The change in the volume at constant pressure is
ΔV= W/P
there fore P = W/ΔV= -3000/-0.01= 3×10^5
Hence the pressure is 
The answer is "heat transfer."
Answer:

Explanation:
Intensity is given by the expresion:

where:
Io = inicial intensity
r1= initial distance
r= final distance


2.B
4.C
3.D
1.C
5.C
Theses are the right answer
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