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2 years ago

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Fill in the blank. Consider the inverse square law: When light leaves a light bulb, it spreads out over more and more space as i

t goes. This makes the light thinner, with less and less light present the further from the light bulb we look. If we stand looking at a light bulb and see how bright it is, then move to be four times farther away from the light bulb, the light from the bulb will look ____________ less bright. Group of answer choices

1 answer:

aliya0001 [1]2 years ago

8 0

Answer:

Explanation:

Intensity of light is inversely proportional to distance from source

I ∝ 1 /r² where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

I₁ /I₂ = r₂² /r₁²

If r₂ = 4r₁ ( given )

I₁ / I₂ = (4r₁ )² / r₁²

= 16 r₁² / r₁²

I₁ / I₂ = 16

I₂ = I₁ / 16

So intensity will become 16 times less bright .

"16 times " is the answer .

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Which of these events is associated with September 11, 2001?

Oxana [17]

Terrorist attacks on the United States is the answer.

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3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

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Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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2 years ago

The position of a ball rolling in a straight line is given by x=2−3.6t+4.4t2, where x is in meters and t in seconds.

kobusy [5.1K]

I'm assuming the question is time it will take for ball to reach ground, if it is then set equation to zero then use the quadratic formula, the possible t value is your answer then

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2 years ago

Is an object speeding up or slowing down of the V final is greater than the V initial?

sertanlavr [38]

Answer:

V is greater

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2 years ago

A mass moves back and forth in simple harmonic motion with amplitude A and period T.

Sever21 [200]

a. 0.5 T

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So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

$1 T : 4 A = t : 2 A$

and solving for t we find

$t=\frac{(1T)(2 A)}{4A}=0.5 T$

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

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we can set the following proportion:

$1 T : 4 A = t : 5 A$

And by solving for t, we find

$t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T$

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2 years ago