Answer:
= 285 Joules
Explanation:
a) answer can be found out in attachment
(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:
p V = nR T to see that the temperature also doubles.
.So... temp for isothermal compression = 355×2 = 710 K
.(c) The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or
max pressure = 2×240000 Pa = 480000 Pa = 4.80 x 10^5 Pa
(d) total work done by the piston = workdone during isothermal compression - work done during expansion =
= nRT ln(V initial / V final)-p (V initial - V final)
= nRT ln(2) - nR(T final - T initial)
= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)
= 285 Joules
Answer:
decibel gain is 10 db
Explanation:
given data
voltage v1 = 5 volts
voltage v2 = 15 volts
to find out
decibel gain
solution
we have given that 5 volt voltage that is increase to 15 volts
so
we know here decibel gain formula that is
decibel gain = 20 log(v2/v1) ....................1
put here value of v1 and v2 in equation 1
decibel gain = 20 log(15/5)
decibel gain = 20 log3
decibel gain = 20 × 0.4771
decibel gain = 9.54 db = approx 10 db
so decibel gain is 10 db
Answer:
f =63.64Hz
Explanation:
The equation of the transverse periodic wave
y = 0.005 sin(20.0x + 2ft)
From the equation we can deduce that
Amplitude A = 0.005
Wave number K = 20
Angular frequency w = 2πf
We are asked to find the frequency (f)
the formular for the wave number is given as K = 2π / lambda
Lambda = 2π/ K
Lamda is the wavelength
= 2 × 3.142 /20
= 0.3142m
The velocity of the wave is expressed as v = frequency × lamba (wavelength)
Frequency = velocity / wavelength
f = 20/ 0.3142
f = 63.65Hz
Answer:
Explanation:
Given that,
Force acting on an object, F = 30 N
Mass of the object, m = 4 kg
It is moving with a constant velocity of 2 m/s across a level surface.
We need to find the coefficient of friction between the object and the surface. Let it is μ. Force in terms of coefficient of friction is given by :
F = μ N, Where N is normal force, N = mg
So, the coefficient of friction between the object and the surface is 0.75.
