During constructive interference two waves travel through the same medium and the effects of the first wave are ___ the effects
1 answer:
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The question is not complete and the complete question is;
An insulated piston-cylinder device contains 4 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-Vdiagram with respect to saturation lines.
Answer:
Voltage of source = 175.33 V
I've attached an image of the P-Vdiagram of the process with respect to saturation lines.
Explanation:
To solve this question, we'll adopt the following assumptions.
- The kinetic and potential energy changes are negligible
-The thermal energy stored in the cylinder is negligible
-The cylinder is well insulated and thus heat transfer is negligible.
If the contents of the cylinder is taken as the system, the energy balance is;
Ein - Eout = ΔEsystem
Thus, [W(e-internal) + W(pw-internal)] - Wout = ΔU
Thus, IVΔt + W(pw-internal) = ΔH
= m(h2 - +1)
Now,looking at steam table A-5 attached to this answer, at P =175, x1 is calculated to be 0; dryness fraction, x2 = 0.5; vf = 0.001057m³/kg and hf = 487.01KJ/Kg and hg at saturation vapour = 2700.2
So, from this question v1=vf = 0.001057m³/kg and h1 = hf = 487.01KJ/Kg.
Also, h2 = hf + x2 (hg - hf) = 487.1 + 0.5(2700. 2 - 487.1) = 1593.65 KJ/Kg
The mass of the water is defined as;
M=V/v1
Volume(V) = 4L from the question. Since v1 is in m³/kg, let's convert V to m³. So V=4/1000 = 0.004
So, M = 0.004/0.001057 = 3.784kg
Now the formula for the voltage is ;
V = [M(h2 - h1) - W(pw-internal)] /(IΔt)
Δt=45minutes. We convert it to seconds to get 45x60 =2700 seconds
V = [[3.784 x (1593.65 - 487.1) - 400]/(8 x2700)] x 1000 = 175.33 V
Answer:
Mg= 1.3418 kg*m/s
Explanation:
Mass of rifle
Recoil velocity of rifle
Momentum of rifle
mass of bullet
mb=7.2x10^{-3} kg=0.0072 kg
velocity of bullet relative to muzzel
v_{b} = 601 m/s
velocity of bullet relative to earth= 601 - 1.95=599.05 m/s
momentum of bullet
the momentum of the propellant gases = momentum of rifle - momentum of bullet
the momentum of the propellant gases
Mg=5.655 -4.3132
Mg= 1.3418 kg*m/s
the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle is 1.3418 kgm/s