A train travels 225 km in 2.5 hours. What is the trains average speed

The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of

Korvikt [17]

Answer:

The mass of the earth, $M=6.023\times 10^{24}\ kg$

Explanation:

It is given that,

Time taken by the moon to orbit the earth, $T=27.3\ days=2358720\ m$

Distance between moon and the earth, $r=384000\ km=384\times 10^6\ m$

We need to find the mass of the Earth using Kepler's third law of motion as :

$T^2=\dfrac{4\pi^2}{GM}r^3$

$M=\dfrac{4\pi^2r^3}{T^2G}$

$M=\dfrac{4\pi^2\times (384\times 10^6)^3}{(2358720)^2\times 6.67\times 10^{-11}}$

$M=6.023\times 10^{24}\ kg$

So, the mass of the earth is $6.023\times 10^{24}\ kg$ . Hence, this is the required solution.

7 0

3 years ago

URGENT!!!! WILL GIVE BRAINLIEST!!!!!! 15 POINTS

Artyom0805 [142]

It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)

3 0

3 years ago

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The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

anastassius [24]

The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

5 0

3 years ago

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Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m

Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

5 0

3 years ago

Briefly explain how a resonance tube works

Lera25 [3.4K]

Answer:

As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency.

4 0

2 years ago

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