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3 years ago

Prove that..please help

Physics

1 answer:

GaryK [48]3 years ago

8 0

$\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}$

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

$\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}$

Also,

$\large{ \longrightarrow{ \rm{ F \propto \dfrac{1}{ {d}^{2} } }}}$

Combining both, We will get

$\large{ \longrightarrow{ \rm{F \propto \dfrac{ m_1 m_2}{ {d}^{2}}}}}$

Or, We can write it as,

$\large{ \longrightarrow{ \rm{F \propto \: G \dfrac{ m_1 m_2}{ {d}^{2} }}}}$

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

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A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

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Answer:

130m

Explanation:

You just have to multiply velocity by the time traveled:

100m/s * 1.3s = 130m!

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3 years ago

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In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

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3 years ago

A beam of light strikes one face of a window pane with an angle of incidence of 30.0°. The index of refraction of the glass is 1

natka813 [3]

Answer:

(a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

Explanation:

Given that,

Angle of incidence = 30.0°

Index of reflection of glass = 1.52

(a). We need to calculate the angle of refraction for the ray inside the glass

Using snell's law

$\dfrac{\sin i}{\sin r}=\mu$

$\sin r=\dfrac{\sin i}{\mu}$

Put the value into the formula

$\sin r=\dfrac{\sin 30}{1.52}$

$r=\sin^{-1}(0.329)$

$r=19.26^{\circ}$

(b). We know that,

The incident ray and emerging ray is equal then the ray will be parallel.

We need to prove that the rays in air on either side of the glass are parallel to each other

Using formula for emerging ray

$\dfrac{\sin e}{\sin r}=\mu$

$\sin e=\sin r\times \mu$

Put the value into the formula

$\sin e=0.3289\times 1.52$

$e=\sin^{-1}(0.499)$

$e=29.9\approx 30^{\circ}$

So, $\sin i=\sin e$

This is proved.

Hence, (a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

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3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

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Prove that..please help​

Prove that..please help