Answer:
The second one
Explanation:
The second one is joining so it is the one which belongs to X, first one belongs to Z
Answer:
f = 276.6 Hz
Explanation:
This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.
In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is
L = λ/ 4
speed is related to wavelength and frequency
v = λ f
λ = v / f
we substitute
L = v / 4f
f = v / 4L
the speed of sound at 20ºC is
v = 343 m / s
let's calculate
f =
f = 276.6 Hz
ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
Answer:
Option B
Explanation:
Looking at the 3 galvanometer readings given above, for galvanometer A, the reading is -2 mA.
For galvanometer B, the reading is 4 mA.
While for galvanometer C, the reading is -5 MA
Thus, option B is correct.