Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
Answer:
Theory
Explanation:
Conservation of energy is explained as a scientific law and not a theory because it does not explain why energy is conserved.
A law is a the statement of a scientific fact. It is a product of repeated experiment and observation through time. Most laws do not explain the reason for the logic behind their premise.
A theory on the other hand provides an explanation for an observed phenomenon. Most theories are no immutable. They are often changed when new finds are reported or made.
Laws are immutable and they stand still.
Answer:
-3m+7m = 4m
Explanation:
As he walks south, he is going down 3m (-3m). Then he walks up 7m (+7m).
You subtract the final position from the initial position to get displacement.
7m - 3m = 4m
Answer:
1. 230 kg...
W = m * g
W= 230 kg * 9.81 m/s^2
<u>W= 2256,3 N</u>
m= 230kg , W = 2256,3 N , g= 9.81 m/s^2
2. 887 N
W= m * g
887 N = m * 9.81 m/s^2
<u>m= 90,42 kg</u>
m= 90,42 kg, W = 887 N , g= 9.81 m/s^2
3. 420 kg
W= m * a
w= 420 kg * 9.81 m/s^2
<u>w=</u><u> </u><u>4120,2 N</u>
m= 420 kg , W = 4120,2 N , g= 9.81 m/s^2
4. Determine the gravity on Pluto where a 15 kg object weighs 55.5N.
w = m * g
55.5 N = 15 kg * g
<u>a= 3,7 m/s^2</u>
m = 15 kg , W= 55.5 N , g= 3,7 m/s^2
This question is a big fat non sequitur !
The wavelength of radio waves traveling through vacuum only depends on the frequency that the radio station is licensed to broadcast on, (which had better be the frequency of the transmitter that they buy and use, or they're in big trouble).
The wavelength does NOT depend on the type of modulation that's used to put information onto the signal.
An amateur radio (ham) operator may very well start out using FM to talk over his radio to somebody else, and then for some reason they may decide to switch to AM. They can do that without ANY change in the wavelength of their transmissions.
Now, in the USA and many other countries, it so happens that all AM stations are licensed by their governments to transmit their programs on a channel somewhere between 500 KHz and 1.6 MHz, and all FM stations are licensed by their governments to transmit their programs on a channel somewhere between 88 MHz and 108 MHz. (And THAT's what the radio receivers in these countries are built to receive.)
Then we might say that all of the AM stations are grouped around 1 MHz, and all of the FM stations are grouped around 100 MHz. The FM frequencies are very roughly 100 times the AM frequencies, so the AM wavelengths are very roughly 100 times the FM wavelengths. That's <em>choice (3)</em> .
But please don't get the idea that it has anything to do with using AM or FM technology. It's just a matter of where in the spectrum the government decided to put the AM stations and where they put the FM stations.
For that matter . . . An analog TV station uses an AM signal for the picture and an FM signal for the sound, and it all goes in the same channel, with just about the same wavelengths !
