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larisa [96]
2 years ago
10

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
Physics
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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A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor
Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field  is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

\Delta A = \frac{1}{2} \Delta \theta R^2

The  formula for the induced emf is

E = \frac{\Delta  \phi}{\Delta  t}

\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

=B\frac{\Delta  A}{\Delta  t}

B is the magnetic field strength

substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

we plug in the values in the equation above

so, the induce EMF will be

E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V

6 0
3 years ago
How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2
soldier1979 [14.2K]
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)

3 0
3 years ago
Read 2 more answers
If you sight Polaris with an astrolabe and determine its altitude is 20° above the horizon, then what does this tell you about y
oksano4ka [1.4K]
If you sight Polaris at 20 degrees above your Northern Horizon then you know that your latitude is 20 degrees north of the equator.
3 0
3 years ago
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A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be
Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

                      F = m \times a

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

              F = m \times a      

                 = 1090 \times (\frac{21}{16})

                 = 1430.625 N

Also, it is known that

      sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}

      sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}

        \theta = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

8 0
3 years ago
What type of heat is transferred to glass casserole dish
madreJ [45]
If i thought about this properly the answer is b
5 0
3 years ago
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