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RUDIKE [14]
3 years ago
8

Which of Kepler's laws states that planets sweep areas in equal times?

Physics
1 answer:
rodikova [14]3 years ago
8 0
Keplers laws states that planets sweep areas in equal times is second
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During anomalous expansion of water ,heat transfer is limited to conduction and radiation only ,explain.​
aalyn [17]

Heat transfer is limited to conduction and radiation only in anomalous expansion of water simply because of the temperature at which the expansion occurs and density

<h3>What is anomalous expansion of water?</h3>

Anomalous expansion of water is a property of water in which water expands instead of contracting.

  • Anomalous expansion of water makes water less dense.
  • The major effect of this anomalous expansion it will still remain less dense and at the surface of water.
  • Interestingly, this expansion occurs when it is cooled from 4°C to 0°C.

Learn more about properties of water:

brainly.com/question/18681949

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2 years ago
Which of these is a good example of Newton's First Law of inertia
Andru [333]

Answer:

  1. one body movement to the side when a car makes a sharp turn
3 0
3 years ago
Read 2 more answers
A vocalist with a bass voice can sing as low as 92 Hz.
Inessa05 [86]

Answer:

  • 3.26 x 10 to the power of 6

Explanation:

c = lambda × frequency

5 0
3 years ago
Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on
Natasha2012 [34]

Answer:

8.8\times 10^{-6} W/m^2

Explanation:

We are given that

Wavelength,\lambda=571 nm=571\times 10^{-9} m

1 nm=10^{-9} m

R=75 cm=\frac{75}{100}=0.75 m

1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

Therefore

\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}

\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2

I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

6 0
3 years ago
The area under the curve of the net external force vs time graph is equal to __________ or ________.
Sophie [7]

Impulse delivered

or

Change in momentum.

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3 years ago
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