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Neko [114]
4 years ago
7

On the 1-to-10-billion scale, about how far is it to the nearest stars besides the sun? on the 1-to-10-billion scale, about how

far is it to the nearest stars besides the sun? 400 kilometers 4 kilometers 1,000 kilometers 10,000 kilometers 4,400 kilometers
Physics
1 answer:
aliina [53]4 years ago
8 0
Not sure what your question means but the nearest star is Alpha Centauri which is about 4.2 light years (ly) away.  This is roughly 4x10¹³ km away.  A billion is 10⁹ so this is 4x10⁴ larger than a billion.  I'd say the last one then...
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Which example best represents translation kinetic energy
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4 years ago
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What simple machines are used to create the shovel?
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A shovel would classify as choice (A) because the head of the shovel is the wedge while the body of the shovel is the lever 
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Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
Nat2105 [25]

Answer:

Explanation:

Given

diameter d=20 \mu m

density \rho =1300 kg/m^3

frequency \nu =150 Hz

Length of silk strand L=14 cm

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\nu =\sqrt{\frac{T}{\mu }}

The expression for Fundamental Frequency

f=\frac{\nu }{2l}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}

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f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}

T=4\rho \cdot A\cdot l^2\cdot f^2

T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2

T=7.2\times 10^{-4} N

8 0
4 years ago
I need help ASAP I need to get this right plz plz plz!!!!!
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<em><u>The</u></em><em><u> </u></em><em><u>atomic</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>consists</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>protons</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u>.</u></em>

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3 0
3 years ago
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Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel, and after 5.5 revolutions,
spin [16.1K]

Answer:

The angle is 23.2 radians, equivalent to 3.69 revolutions.

Explanation:

First, we need to find the angular acceleration of the wheel. This can be done using one of the kinematic formulas:

\omega^{2}=\omega_0^{2}+2\alpha\theta\\\\\implies \alpha=\frac{\omega^{2}-\omega_0^{2}}{2\theta}

Since the final angular velocity is zero after 5.5 revolutions (equivalent to 11π radians) we have that:

\alpha=\frac{-(3.15rad/s)^{2}}{2(11\pi rad)}\\\\\alpha=-0.144rad/s^{2}

Now, using the same equation, we can solve for the requested angle:

\theta=\frac{\omega^{2}-\omega_0^{2}}{2\alpha}\\\\\theta=\frac{(1.80rad/s)^{2}-(3.15rad/s)^{2}}{2(-0.144rad/s^{2})}\\\\\theta=23.2rad

Finally, it means that the angle through which the wheel has turned when the angular speed reaches 1.80 rad/s is 23.2 radians, equivalent to 3.69 revolutions.

8 0
3 years ago
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