Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 × 
m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 = 
Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 = + 1000 × 9.8 ×36
pressure at bottom P2 =4.52 × Pa
so from gas law
here p is pressure and v is volume and t is temperature
so put here value and find v1
V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³
Answer:
1.68 s
Explanation:
From newton's equation of motion,
a = (v-u)/t.................................. Equation 1
Making t the subject of the equation
t =(v-u)g............................. Equation 2
Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.
Note: Taking upward to be negative and down ward to be positive,
Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²
t = (0-8.20)/-9.8
t = -8.20/-9.8
t = 0.84 s.
But,
T = 2t
Where T = time taken for the bowling pin to return to the juggler's hand.
T = 2(0.84)
T = 1.68 s.
T = 1.68 s
Answer:
methyl orange, methyl red,phenoptalin, merhy red
Explanation:
all this following are indicators use to check the end point of a reaction
Answer:
Learning the formula.multiply mass accelebrations.the force(F)required to move an object of mass(M) with an acceleration (a) is given by the formula F = m x a.so, force = mass multiplied by accelebration.
The total circuit current at the resonant frequency is 0.61 amps
What is a LC Circuit?
- A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
- These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.
Q =15 = (wL)/R
wL = 30 ohms = Xl
R = 2 ohms
Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance
| Zs | = 30.07 <86.2° ohms
Xc = 1/(wC) = 30 ohms
The impedance of the LC circuit is found from:
Zp = (Zs)(-jXc)/( Zs -jXc)
Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°
I capacitor = 277/-j30 = j9.23 amps
I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps
I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps
Hence, the total circuit current at the resonant frequency is 0.61 amps
To learn more about LC Circuit from the given link
brainly.com/question/29383434
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