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3 years ago

What might cause a ballon blown up at a low altitude to burst if it was taken to a higher altitude?

2 answers:

6 0

When you are at a low altitude their is more pressure. If you are at a high altitude, there would be less pressure, making it pop

maks197457 [2]3 years ago

5 0

If a balloon is high up it will pop because pressure here think of it like this when you swim and go under water lets say 12ft you will notice that your head will hurt and it will take your oxygen away faster just like a balloon it has air in it so when it goes to high it will pop because its made of weak material so it will pop. and a low balloon because it will hit things and like i said balloons pop easy it could also get soft rubber because it is not moving and if like the sun was on it than it would pop as well i really hope this helped you

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Which of the following is true for a parallel circuit?

svet-max [94.6K]

Answer:

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3 years ago

What product is obtained from the aldol condensation of cyclohexanone?

Evgesh-ka [11]

Answer:

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-Second product is FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Explanation:

Please see the attached image for complete chemical reaction of aldol condensation of cyclohexanone

Aldol Condensation is a form of electrophilic substitution reaction in which the alpha carbon in enols or enolate anions is substituted by an electrophile to form carbon-carbon bond. Cyclohexanone also known as the first ketone consists of two alpha-carbons and four potential substitutions i.e alpha-hydrogens but none of the hydrogen on the ring is substituted. Ketones such as cyclohexanone are much more acidic than their parent hydrocarbon.

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-cyclohexanone that further undergoes dehydration resulting into FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Based on the explanations above, the compound formed is shown in the image.

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3 years ago

What law describes light

Darina [25.2K]

The law of reflection

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3 years ago

What is an example of a wave that is not mechanical and how is it different?

vovikov84 [41]

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3 years ago

The radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm?

marin [14]

Using r to represent the radius and t for time, you can write the first rate as:

drdt=4mms

r=r(t)=4t

The formula for a solid sphere's volume is:

V=V(r)=43πr3

When you take the derivative of both sides with respect to time...

dVdt=43π(3r2)(drdt)

...remember the Chain Rule for implicit differentiation. The general format for this is:

dV(r)dt=dV(r)dr(t)⋅dr(t)dt with V=V(r) and r=r(t) .

So, when you take the derivative of the volume, it is with respect to its variable r (dV(r)dr(t)) , but we want to do it with respect to t (dV(r)dt) . Since r=r(t) and r(t) is implicitly a function of t , to make the equality work, you have to multiply by the derivative of the function r(t) with respect to t (dr(t)dt) as well. That way, you're taking a derivative along a chain of functions, so to speak (V→r→t ).

Now what you can do is simply plug in what r is (note you were given diameter) and what drdt is, because dVdt describes the rate of change of the volume over time, of a sphere.

dVdt=43π(3(20mm)2)(4mms) =6400πmm3s

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

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3 years ago