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klemol [59]
3 years ago
5

Explain why water boiling in a container stopsboiling momentarily when the lid is removed​

Physics
1 answer:
Dmitriy789 [7]3 years ago
8 0

Answer:

It's because it tips over the threshold from nucleate boiling, which we can see, to convection boiling, which we can't. ... Even if the steam stayed in the pot, it would still stop boiling when you removed the heat. The steam and water in a liquid/vapour mixture are at the same temperature (100ºC).

Explanation:

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An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only
topjm [15]

Answer:

Explanation:

Let the potential difference between the middle point and one of the plate be ΔV .

electric potential energy will be lost and it will be converted into kinetic energy .

Electrical potential energy lost = Vq , where q is charge on charge particle .

For proton

ΔV× q = 1/2 M V² ( kinetic energy of proton )

where M is mass and V be final velocity of proton .

For electron

ΔV× q = 1/2 m v² ( kinetic energy of electron  )

where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .

As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same

1/2 M V² =  1/2 m v²

(V / v )² = ( m / M )

(V / v ) = √ ( m / M )

In other words , their velocities  are  inversely proportional to square root of their masses .

4 0
3 years ago
A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is
Klio2033 [76]

Answer:

   θ = 10.28º

Explanation:

To find the angle of refraction use the equation of refraction

        n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for incident light and index 2 is for refracted light.

         sin θ₂ = n₁ / n₂ sin θ

let's calculate

         sin = 1 / 1.3 sin 0.23

         sin = 0.175

        θ= 0.17528 rad

let's reduce to degrees

       θ = 0.17528 rad (180ª / pi rad)

       θ = 10.28º

6 0
3 years ago
What is the range of motion of the elbow if extension is 0° and flexion is 145°?
inna [77]

Answer:

0 to 145 degrees

Explanation:

The normal range of flexion and extension is from 0 to 145 degrees.

6 0
2 years ago
Graphs do not need to have a title.true or false
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The answer for this question should be "false".
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3 years ago
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Which of the following is an example of the wind slowly causing changes in Earth's landforms?
stepan [7]
It is called Wind Erosion.
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3 years ago
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