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valentina_108 [34]
2 years ago
10

A 1. 18 kg gold cube hangs at the end of a 4. 00 m long string. Rhogold = 19. 3 × 103 kg/m3; rhomercury = 13. 6 × 103 kg/m3. Whe

n the gold cube is immersed in mercury, the tension in the string, in n, is
Physics
1 answer:
VashaNatasha [74]2 years ago
6 0

When the gold cube is immersed in mercury, the tension in the string in Newtons is 3.142N.

<h3>What is tension?</h3>

Tension is the force acting on the linear object like string, chain or rope due to pulling.

Volume of gold V = mass / density

V = 1.18 /19.3x 10³ =61.1 x 10⁻⁶ m³

Tension in the string after immersing will be

T = [ρ(Gold)  -ρ(Hg)] g. V

T =[ 19.3x 10³ - 13.6 x 10³] x 9.81 x 61.1 x 10⁻⁶

T =3.416 N
Thus, the tension in the string is 3.42 N.

Learn more about tension.

brainly.com/question/4087119

#SPJ4

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A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
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This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

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Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

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Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

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