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Dafna1 [17]
3 years ago
14

You live on a street that runs East to West. You just had 2 inche of snow and you live on the North side of the street. You retu

rn from class at 2PM and notice all the snow on your sidewalk is gone but across the street it is still there. No one removed the snow. How did it go away?
Engineering
1 answer:
RSB [31]3 years ago
5 0

Answer:

The heat from the sun melted it

Explanation:

If the street runs east to west, houses on the south (across the street) will project shadows on their sidewalk, while the northern sidewalk will be illuminated. This is for the northern hemisphere, on the southern hemisphere it would be the other way around.

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can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.​
marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

6 0
3 years ago
A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside
djverab [1.8K]

Answer:

The diameter increases

Explanation:

The expansion in the metal is uniform in every dimension

4 0
2 years ago
What is the focus of 7th grade civics
Jobisdone [24]
C. seems like the best answer. i may be wrong so don’t quote me on that
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3 years ago
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The complex power of a load is 10-10j VA. What component should be added in parallel with the load so that the new load has a un
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7 0
3 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
3 years ago
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