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3 years ago

9

Two objects that may be considered point masses are initially separated by a distance d. The separation distance is then decreas

ed to d/4. How does the gravitational force between these two objects change as a result of the decrease?

1 answer:

ozzi3 years ago

8 0

Answer:

Increased by 16 times

Explanation:

F = Gravitational force between two bodies

G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/kg s²

m₁ = Mass of one body

m₂ = Mass of other body

d = distance between the two bodies

$F=\frac{Gm_1m_2}{d^2}\\ F=\frac{1}{d^2}\quad \text {(as G and masses are constant)}$

$F_{new}=\frac{1}{\left (\frac{d}{4}\right )^2}\\\Rightarrow F_{new}=\frac{1}{\frac{d^2}{16}}\\\Rightarrow F_{new}={16}\times \frac{1}{d^2}\\\Rightarrow F_{new}=16\times F$

∴Force will increase 16 times

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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

a cylinder of mass 34.5 kg rolls without slipping on a horizontal surface. At a certain instant, its center of mass has a speed

Stolb23 [73]

Answer:

778 J

Explanation:

Rotational energy is:

RE = ½ Iω²

For a solid cylinder I = ½ mr².

Rolling without slipping means ω = v/r.

RE = ½ (½ mr²) (v/r)²

RE = ¼ mv²

Plug in values:

RE = ¼ (34.5 kg) (9.5 m/s)²

RE ≈ 778 J

Round as needed.

4 0

4 years ago

Read 2 more answers

I don’t get this question please help me

zvonat [6]

oof, that question is difficult! i would say its d because if you pull a slingshot farther than another one, it would have more potential energy.

6 0

3 years ago

Read 2 more answers

A ski jumper travels down a slope and leaves

Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

$v_x = 28 m/s$

and the distance covered along the horizontal direction in a time t is

$d_x = v_x t$

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity $u_y = 0$ and constant acceleration $g=9.8 m/s^2$ (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

$d_y = \frac{1}{2}gt^2$

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

$\theta = 36^{\circ}$ below the horizontal

This happens when:

$tan \theta = \frac{d_y}{d_x}$

Substituting and solving for t, we find:

$tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s$

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

$d_x = v_x t = (28)(4.15)=116.2 m$

$d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m$

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m$

7 0

3 years ago

When you push on an object such as a wrench, a steel pry bar, or even the outer edge of a door, you produce a torque equal to th

Korvikt [17]

<span>Since youc oncetrate all your force directly towards the moment arm it means that you push it at an angle of your force is directed to the left or the right and I bet that it must be 90</span> degrees to the bar. Obviuosly, if you are about to push it you will do it straight up but not in a zig zag way. In other words, it should be perpendicular to the arm because the<span> torque can be produced only if force is applied at a constant index (90).
Hope that helps! Regards.</span>

5 0

3 years ago