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3 years ago

8

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

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:

one body movement to the side when a car makes a sharp turn

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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

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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

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

$\phi=7.74 rad$

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

$\beta=5.3 rad$

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

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

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

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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