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3 years ago

The air in a cool region underneath cloud cover will have _____ a region with no cloud cover.

Physics

2 answers:

Slav-nsk [51]3 years ago

7 0

<u>Answer:</u>

The air is a <u>homogeneous mixture of gases</u> that make up the Earth's atmosphere, which remain around the planet by the attraction force of gravity.

Its properties changes with the altitude, the pressure, the temperature and the density. For example, there is a relation that is <u>inversely proportional</u> between the temperature and the air density.

<h2>When the temperature decreases the density increases. </h2>

Therefore, the air in a cool region of the atmosphere will be denser than the air in a hot region of the atmosphere.

In other words:

<h2>The air in a cool region underneath cloud cover will have <em><u>a higher density than</u></em> a region with no cloud cover </h2><h2> </h2>

Aleksandr-060686 [28]3 years ago

5 0

The air in a cool region underneath cloud cover will have a higher density than a region with no cloud cover. A region with no cloud cover has warmer air than the surrounding air, it becomes less dense and begins to rise, drawing more air in underneath.

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Explain why frog will not look green under the red light?

IrinaVladis [17]

A frog can be many different colours. It appears green under normal 'white' light because it absorbs all the other colours in the light's spectrum apart from green. It reflects the green light back and that is picked up by your eye.

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3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

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3 years ago

A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l

MrRissso [65]

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

$F = -kx$

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

$F=mg$

Inserting this into Hooke's Law

$mg=-kx$

computing it for x,

$-x=mg/k$

This is the model which will tell the length of the spring against change in the mass located in the pan.

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3 years ago

Newton’s laws do not apply to small objects?

Soloha48 [4]

Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

when body at rest always want to be at rest

or body at motion always want to be at motion

unles an external force acts upon it

for example a eraser on the table will be at rest

if so e apply some force then it comes motion

so, Newton s law apply to small object s

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3 years ago