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coldgirl [10]
2 years ago
9

After reading through the code, what will happen when you click run?* 1 point when run move forward while there is a pile do rem

ove 1 O Nothing will happen. O The farmer will dig a hole and add to the pile of dirt. O The farmer will never stop removing dirt. O The farmer will remove dirt as long as there is a pile, then stop when the pile is done. 1 point After reading through the code, what will happen when you click run? * when an​

Engineering
1 answer:
KATRIN_1 [288]2 years ago
5 0

Explanation:

I'm not exactly a master at coding, but I'm pretty sure that:

The farmer will remove dirt as long as there is a pile, then stop when the pile is done.

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If a vacuum gau ge reads 9.62 psi, it means that: a. the very highest column of mercury it could support would be 19.58 inches.
scZoUnD [109]

Answer:All of the above

Explanation:

9.62 psi means 497.49 mm of Hg pressure

for (a)19.58 inches is equals to 497.49 mm of Hg

(b)atmospheric pressure is 14.69 psi

vaccum gauge is 9.62psi

absolute pressure is=14.69-9.62=5.07

(c)vaccum means air is sucked and there is negative pressure so it tells about below atmospheric pressure.

thus all are correct

8 0
3 years ago
A PLL is set up so that its VCO free-runs at 8.9 MHz. The VCO does not change frequency unless its input is within plus or minus
pickupchik [31]
It’s A 75khz because it’s plus or minus so if u add it would be too much
5 0
2 years ago
I'll mark brainliest plz help
Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0
3 years ago
How many 10" diameter circles can be cut from a semicircular shape that has a 20"
N76 [4]

9514 1404 393

Answer:

  1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

3 0
2 years ago
Read 2 more answers
The primary heat transfer mechanism that quickly warms my hand if I hold it directly above a campfire is: a)-Radiation b)-Induct
Tanzania [10]

Answer:

The correct answer is option 'c':Convection.

Explanation:

When we ignite a campfire the heat produced by combustion heats the air above the fire. As we know that if a gases gains heat it expands thus it's density decreases and hence it rises, if we hold our hands directly above the fire this rising hot air comes in contact with our hands thus warming them.

The situation is different if we are at some distance from the campfire laterally. Since the rising air cannot move laterally the only means the heat of the fire reaches our body is radiation.

But in the given situation the correct answer is convection.  

6 0
3 years ago
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