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elena-s [515]
2 years ago
7

What kinds of natural phenomena does the inverse-square law apply to?

Physics
2 answers:
lesantik [10]2 years ago
7 0

Answer:

Being strictly geometric in its origin, the inverse square law applies to diverse phenomena. Point sources of gravitational force, electric field, light, sound, and radiation obey the inverse square law.

Explanation:

uysha [10]2 years ago
4 0

Answer:

It applies to diverse phenomena

Explanation:

not sure if u need one, just ask if you do :D

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grandymaker [24]
Light because they need light to grow

3 0
3 years ago
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A wire is used as a heating element that has a resistance that is fairly independent of its temperature within its operating ran
dedylja [7]

Answer:

Double the current

Explanation:

The energy delivered by the heater is related to the current by the following relation:

E= I^{2}R t

let R * t = k ( ∴ R and t both are constant)

so E= k I^{2}

Now let:

E2= k I₂^2

E2= 4E

⇒ k I₂^2= 4* k I^{2}

Cancel same terms on both sides.

I₂^2= 4* I^{2}

taking square-root on both sides.

√I₂^2 = √4* I^2

⇒I₂= 2I

If we double the current the energy delivered each minute be 4E.

3 0
2 years ago
Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? S
Zepler [3.9K]

Answer:

About two kilometers away

\rm distance=2.401\ km

Explanation:

Given:

The time gap between the light and sound to travel to the house, t=7\ s

<em>Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.</em>

<u>∴Distance between the lightning-strike and the house:</u>

\rm distance=v\times t

we have the speed of sound as: v=343\ m.s^{-1}

So,

\rm distance=343\times 7

\rm distance=2401\ m

\rm distance=2.401\ km

6 0
2 years ago
One disadvantage of using proprietary licensed software is that
Vaselesa [24]

Answer:

its b

Explanation:

5 0
2 years ago
Read 2 more answers
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
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