All categories

2 years ago

a metal coin has certain properties that can be measured.which property of a coin is different on the moon that is on earth?

Engineering

1 answer:

Sloan [31]2 years ago

8 0

Answer:

Coins weigh less on the Moon.

Explanation:

Gravity is only 1/6th as strong on the Moon than it is on Earth. Where a nickle is about 5 grams on Earth, it is less than 1 gram on the Moon. Gravity is affected by the size of the planet or moon. The Moon is much less massive than the Earth.

You might be interested in

A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. Obtain the yield strength and ultimate str

Charra [1.4K]

Answer:

yield strength before cold work = 370 MPa

yield strength after cold work = 437.87 MPa

ultimate strength before cold work = 440 MPa

ultimate strength after cold work = 550 MPa

Explanation:

given data

AISI 1018 steel

cold work factor W = 20% = 0.20

to find out

yield strength and ultimate strength before and after the cold-work operation

solution

we know the properties of AISI 1018 steel is

yield strength σy = 370 MPa

ultimate tensile strength σu = 440 MPa

strength coefficient K = 600 MPa

strain hardness n = 0.21

so true strain is here ∈ = $ln\frac{1}{1-0.2}$ = 0.223

yield strength after cold is

yield strength = $K \varepsilon ^n$

yield strength = $600*0.223^{0.21)$

yield strength after cold work = 437.87 MPa

and

ultimate strength after cold work is

ultimate strength = $\frac{\sigma u}{1-W}$

ultimate strength = $\frac{440}{1-0.2}$

ultimate strength after cold work = 550 MPa

8 0

2 years ago

Pointssss 100 and brainliest :)

Delvig [45]

Answer:

thank you for the free point have a great rest of your day

7 0

2 years ago

The chart shows the bids provided by four engineers to test a prototype.

klasskru [66]

Answer:

Explanation:

To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete. You have to multiply them to get the total cost of the project.

$\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]$

As you can see, Greg is the least cost-effective because he charges the most for the project.

8 0

3 years ago

A company, studying the relation between job satisfaction and length of service of employees, classified employees into three le

Wewaii [24]

Answer:

Below see details

Explanation:

A) It is attached. Please see the picture

B) First to calculate the overall mean,

μ=65∗25/75+80∗25/75+95∗25/75

μ=65∗25/75+80∗25/75+95∗25/75 = 80

Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634

And E(MSE) = σ^2= 9

C) Yes, it is substantially large than E(MSE) in this case.

D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.

3 0

2 years ago

A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = $\frac{60+180}{2}$ = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0

2 years ago