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kotykmax [81]
3 years ago
10

If Fg=mg solve for g

Physics
1 answer:
Bas_tet [7]3 years ago
6 0

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

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a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t
Oksi-84 [34.3K]

Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L  . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

L = 25 ( 1 + 12 x 10⁻⁶ Δt )

For brass ring

L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

1.004( 1 + 12 x 10⁻⁶ Δt ) =  ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt  =   1 + 20 x 10⁻⁶ Δt

.004 = 7.9518 x 10⁻⁶ Δt

Δt  = 4000 / 7.9518

= 503⁰C.

final temp = 503 + 15 = 518⁰C  .

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3 years ago
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A cheetah runs at a speed of 27.6 m/s. If the
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Answer:

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Explanation:

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3 years ago
Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.
galina1969 [7]

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}

Here, k is the Coulomb's constant whose value is

k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the magnitude of charge will be

\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

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1 year ago
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