All categories

2 years ago

An engine jack is used to lift a 265-lb engine 6'. How much work, in ft-lb, is performed?

Engineering

1 answer:

Vlad [161]2 years ago

5 0

(-1)*exp(+i pi) = 1

formatted cleanly

2x2x2x2x2

You might be interested in

8. What is the purpose of the 300 Log?

Gnom [1K]

Answer:

“An OSHA 300 log is where companies record the injuries that occur at the workplace,” said Luna. “By law, they have to report all the injuries to OSHA.” The OSHA law gives workers and their unions the right to have access to injury logs.

4 0

2 years ago

A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str

just olya [345]

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is $\frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}$

The shear strength= $0.58\sigma_y$ and in this case $\sigma_y=36 000 psi$

Shear strength= $\frac {Load}{Shear area}$ hence making load the subject then

Load=Shear area X Shear strength

Load= $\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi$

3 0

3 years ago

An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [

crimeas [40]

Answer:

Modulus of resilience will be $3216942.308j/m^3$

Explanation:

We have given yield strength $\sigma _y=818MPa$

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience $=\frac{\sigma _y^2}{2E}$ , here $\sigma _y$ is yield strength and E is elastic modulus

Modulus of resilience $=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3$

5 0

3 years ago

A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of

viktelen [127]

Answer:

second-law efficiency = 62.42 %

Explanation:

given data

temperature T1 = 1200°C = 1473 K

temperature T2 = 20°C = 293 K

thermal efficiency η = 50 percent

solution

as we know that thermal efficiency of reversible heat engine between same temp reservoir

so here

efficiency ( reversible ) η1 = 1 - $\frac{T2}{T1}$ ............1

efficiency ( reversible ) η1 = 1 - $\frac{293}{1473}$

so efficiency ( reversible ) η1 = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = $\frac{thernal\ efficiency}{0.801}$

second-law efficiency = $\frac{50}{0.801}$

second-law efficiency = 62.42 %

3 0

3 years ago

Tanya Pierce, President and owner of Florida Now Real Estate is seeking your assistance in designing a database for her business

const2013 [10]

Answer:

the answer is attributes for each entity

5 0

3 years ago