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3 years ago

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An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, conver

ts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R. Are these measurements reasonable? Why?

1 answer:

lesya [120]3 years ago

3 0

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

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In addition to passing an ASE certification test, automotive technicians must have __________ year(s) of on the job training or

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Read 2 more answers

What is an isentropic process?

snow_tiger [21]

Answer: Isentropic process is the process in fluids which have a constant entropy.

Explanation: The isentropic process is considered as the ideal thermodynamical process and has both adiabatic as well as reversible processes in internal form.This process supports no transfer of heat and no transformation of matter .The entropy of the provided mass also remains unchanged or consistent.These processes are usually carried out on material on the efficient device.

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3 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

so

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

$\epsilon = 0.22)$

therefore eccentrcity of orbit is 0.22

6 0

3 years ago

For a bronze alloy, the stress at which plastic deformation begins is 266 MPa and the modulus of elasticity is105 GPa.

pentagon [3]

Answer:

88750 N

Explanation:

given data:

plastic deformation σy=266 MPa=266*10^6 N/m^2

cross-sectional area Ao=333 mm^2=333*10^-6 m^2

solution:

To determine the maximum load that can be applied without

plastic deformation (Fy).

Fy=σy*Ao

=88750 N

7 0

2 years ago