Answer:
a)W= - 720 J
b)ΔU= 330 J
Explanation:
Given that
P = 0.8 atm
We know that 1 atm = 100 KPa
P = 80 KPa
V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)
V₂ = 3 L = 0.003 m³
Q= - 390 J ( heat is leaving from the system )
We know that work done by gas given as
W = P (V₂ -V₁ )
W= 80 x ( 0.003 - 0.012 ) KJ
W= - 0.72 KJ
W= - 720 J ( Negative sign indicates work done on the gas)
From first law of thermodynamics
Q = W + ΔU
ΔU=Change in the internal energy
Now by putting the values
- 390 = - 720 + ΔU
ΔU= 720 - 390 J
ΔU= 330 J
We want to find the combined volume of 3 tennis balls. We will get that the combined volume is 493.7 cm^3
First, remember that for a sphere of diameter D, the volume is:

Where 3.14 is pi.
Here we know that the average diameter of a tennis ball is 6.8cm, then we can replace that in the above equation to find the volume (in average) of a single tennis ball:

Now, in 3 balls of tennis, the combined volume will be 3 times the above one, this is:

If you want to learn more about volumes, you can read:
brainly.com/question/10171109
Answer:
720 V
Explanation:
Given that,
The number of turns in primary coil, N₁ = 60
The number of turns in secondary coil, N₂ = 360
The input rms voltage, V₁ = 120 V
We need to find the output rms voltage of the secondary coil
. The relation between number of turns in primary coil - secondary coil to the input rms voltage to the output rms voltage is given by :

<h3>So, the output rms voltage of the secondary coil is 720 V. Hence, the correct option is (b).</h3>
The bouyancy force is:
Since the wood-lead system is completely submerged, the bouyancy force
is FB = ĎwgVl + ĎwgVb, where Ďw is the density of water,Vl
is the volume of
the piece of lead and Vb is the volume of the wooden block. The weight of the
combined lead and wooden block is: W = ĎlgVl + ĎbgVb. Since the system is
in equilibrium, the bouyancy force must be equal to the total weight:
ĎwgVl + ĎwgVb = ĎlgVl + ĎbgVb
now we can solve for the volume of lead:
ĎwgVl â’ ĎlgVl = ĎbgVb â’ ĎwgVb
Vl(Ďw â’ Ďl) = Vb(Ďb â’ Ďw)
Vl =
Ďbâ’Ďw
Ďwâ’Ďl
Vb
Now we substitute the values for the density of lead Ďl = 11.3 Ă— 103kg/m3 ,
the density of the wood and the density of water Ďw = 1000kg/m3
. We get:
Vl =
600â’1000
1000â’11300
(0.6m Ă— 0.25m Ă— 0.08m) = 4.66 Ă— 10â’4m3