Answer:
0.435atm
Explanation:
cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.185 m3 at a pressure of 0.740 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.315 m3. If the temperature remains constant, what is the final value of the pressure?
Given
Initial pressure P1= 0.740atm
Initial volume V1= 0.185 m3
Final pressure P2= ?
Final volume V2= 0.315 m3
At constant temperature, the pressure of a syste is inversely proportional to volume, by Boyles law then
P1V1=P2V2
P2=P1V1/V2
=(0.185*0.740)/0.315
0.1369/0.315
= 0.435atm
Therefore, final pressure is 0.435atm
Answer:
0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli
1.1m to 0.04125 m
Explanation:
T= 1/f=
if f= 300Hz then T = 1/300 =0.003333 s
if f= 8000 then T= 1/8000 = 0.000125s
now v=f×wave length
or wavelength = speed/ frequency
when f = 300 Hz
wavelength = 330/300=1.1 m
wavelength = 330/8000 = 0.04125m
note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s
Given Information:
Resistance = R = 14 Ω
Inductance = L = 2.3 H
voltage = V = 100 V
time = t = 0.13 s
Required Information:
(a) energy is being stored in the magnetic field
(b) thermal energy is appearing in the resistance
(c) energy is being delivered by the battery?
Answer:
(a) energy is being stored in the magnetic field ≈ 219 watts
(b) thermal energy is appearing in the resistance ≈ 267 watts
(c) energy is being delivered by the battery ≈ 481 watts
Explanation:
The energy stored in the inductor is given by

The rate at which the energy is being stored in the inductor is given by

The current through the RL circuit is given by

Where τ is the the time constant and is given by


Therefore, eq. 1 becomes

At t = 0.13 seconds

(b) thermal energy is appearing in the resistance
The thermal energy is given by

(c) energy is being delivered by the battery?
The energy delivered by battery is

Answer:
For example, a wave with a time period of 2 seconds has a frequency of
1 ÷ 2 = 0.5 Hz.
Explanation: