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Likurg_2 [28]
3 years ago
13

Describe how the boiling point of water on top of a mountain would be different from its boiling point at sea level.

Physics
1 answer:
Sonbull [250]3 years ago
7 0

Answer:

At elevated altitudes, any cooking that involves boiling or steaming generally requires compensation for lower temperatures because the boiling point of water is lower at higher altitudes due to the decreased atmospheric pressure.

Explanation:

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How many atoms of Pb are there in the compound 3Pbl2
mash [69]

Answer: there are 6 atoms.

Explanation: 3 parts of Pb times 2 atoms per part = 6

good luck:)

5 0
4 years ago
Read 2 more answers
A softball player moving 3.89 m/s
Ahat [919]

Answer:

0.119 s

Explanation:

Given that

U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m

Lets take final speed of the softball after covering 4.8 m = V

We know that

V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s

Also We know that

V=U+at\\

Putting the value of V ,U and a in the previous equation  We get

3.7=3.89-1.44\times t\\t=0.119\ s

Therefore slide time will be 0.119 s

3 0
4 years ago
When the wedges are removed, the cars will move. Predict which direction they will move and when they will stop
Neko [114]

Answer: hello your question lacks the required diagram attached below is the required diagram

answer : Both cars will move backwards and stop due to friction.

Explanation:

Given that both cars are negatively charged, When the wedges are removed both cars will move backwards ( repelling each other ) because they are like poles, and Like poles repel each other. while unlike poles attract each other ( forward movement )  .

The cars will later come to a stop due to frictional forces between the cars and the surface.

6 0
3 years ago
"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3
avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

We know that magnetic field due to long wire is given by B=\frac{\mu _0i}{2\pi r}

It is given that magnetic field is zero at some distance so

\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}

Here i_1=3\ A \ and\  i_2=8\ A

So \frac{3}{x}=\frac{8}{24-x}=6.5454\ m

3 0
3 years ago
You drop a stone down a well that is 19.60 m deep. How long is it before you hear the splash? The speed of sound in air is 343 m
ki77a [65]

So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

\boxed{\sf{\bold{t = \frac{s}{v}}}}

With the following condition :

  • t = interval of the time (s)
  • s = shift or displacement (m)
  • v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

  • h = height or any other displacement at vertical line = 19.6 m
  • g = acceleration of the gravity = 9.8 m/s²
  • v = velocity = 343 m/s

What was asked :

  • \sf{\sum t} = ... s

Step by step :

  • Find the time when the object falls freely until it hits the water. Save value as \sf{\bold{t_1}}

\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}

\sf{t_1 = \sqrt{4}}

\sf{\bold{t_1 = 2 \: s}}

  • Find the time when the sound propagate through air. Save value as \sf{\bold{t_2}}

\sf{t_2 = \frac{h}{v}}

\sf{t_2 = \frac{19.6}{343}}

\sf{\bold{t_2 \approx 0.06 \: s}}

  • Find the total time \sf{\bold{\sum t}}

\sf{\sum t = t_1 + t_2}

\sf{\sum t \approx 2 + 0.06}

\boxed{\sf{\sum t \approx 2.06}}

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

3 0
3 years ago
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