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3 years ago

According to Newton’s Law of gravity, which of the following is inversely related to the square of the distance from earth?

Physics

2 answers:

Dima020 [189]3 years ago

3 0

I think it’s D

I hope this helps ❤️

Travka [436]3 years ago

3 0

Answer:

B. Weight

Explanation:

If you look at my crude drawing of two bodies attracted by each other (hey that's gravity, not love!) and to Newton's law of gravity:

$F_{1} =F_{2} =G\frac{m_{1} m_{2} }{r^{2} }$

For this to work we have to assume that the Forces are equal and the whole system is in equilibrium.

Now if we take a look at the picture, r is the distance between the two bodies, G is the gravitational constant, and m is the mass either of body 1 or 2, so, if we increase the distance, the force of gravity will decrease by the square root of that unit, because it is the denominator of the fraction.

Now emphasis on "Force of Gravity" Force is the only factor inversely proportionate to the square root of the distance, and the only answer that is actually a Force from the options is weight (m*g).

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

What is an easy way to balance chemical equations?

Lorico [155]

write out what you have on both sides, then just use basic multiplication to try and even out both sides. I can help if you need me to balance some for you!!

8 0

3 years ago

-A 180 kg hippo is riding a bicycle at a speed of 6.0

Vlad1618 [11]

Answer:

0.0675 seconds

Explanation:

From the question,

We apply newton's second law of motion

F = m(v-u)/t.................... Equation 1

Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.

make t the subject of the equation

t = m(v-u)/F................... Equation 2

Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)

Substitute these value into equation 2

t = 180(0-6.0)/-1600

t = -1080/-1600

t = 0.0675 seconds.

8 0

3 years ago

Lus

patriot [66]

Answer:

hey but the person at the top is right

7 0

3 years ago

Electromagnetic radiation is:

ruslelena [56]

Hello there.

<span>Electromagnetic radiation is:

</span><span>A- energy that is electric and magnetic
</span>

5 0

3 years ago