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NNADVOKAT
2 months ago
14

Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried

several different techniques to do this. One method was to cut pivot holes into the ends of the stone and then use oxen to pull the column. The 4-ft diameter column weighs 13200 lbs, and the team of oxen generates a constant pull force of 1500 lbs on the center of the cylinder G. Knowing that the column starts from rest and rolls without slipping, determine the velocity of its center G after it has moved 5 ft and the minimum static coefficient of friction that will keep it from slipping.

Physics
1 answer:
Yuliya22 [3.3K]2 months ago
7 0
This should cover all your needs:)
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A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Softa [3030]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Reservoir pressure = 10 atm

T_1 = Reservoir temperature = 300 K

P_2 = Exit pressure = 1 atm

T_2 = Exit temperature

R_s = Specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

Assuming isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

Flow density at exit measures 2.2721 kg/m³

4 0
4 months ago
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect
Softa [3030]

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

Thus,

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now, by applying integration to the equation above

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

4 0
3 months ago
Layla is researching one of the outer planets. So far, she knows only that it is either a gas giant or an ice giant. Which two s
Keith_Richards [3271]

E) This planet has a greater diameter than any of the four terrestrial planets that orbit close to the Sun.

B) A substantial amount of rocky material accumulated to create a planet larger than Earth.

(Which two statements are the most accurate)

- A, C, and D simply do not make sense.

7 0
3 months ago
Read 2 more answers
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