What method would produce disease resistant cattle in the shortest amount of time?

An rv is traveling 60 km/h along a highway at night. a boy sitting near the driver of the rv turns a flashlight on and shines it

nika2105 [10]

Speed of light relative to the boys = 300,000,000 m/s

Speed of light relative to a stationary observer on the side of the road =

<h3>300,000,000 m/s</h3><h3>What is special theory of relativity?</h3>

The theory of special relativity explains how speed affects space, time, and mass.
Small amounts of mass (m) can be interchangeable with large amounts of energy (E), as defined by the classic equation E = mc2, according to the theory, which offers a means for the speed of light to define the link between energy and matter.

<h3>What are the special relativity's guiding principles?</h3>

The laws of physics should be independent of the uniform motion of an inertial frame of reference, and the speed of light should be constant in any such frame, according to two fundamental tenets that constitute the basis of the special theory of relativity.

To learn more about special relativity visit:

brainly.com/question/12497537

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3 0

2 years ago

hey my friend rly needed help and i wish i could help do you guys kn what .1895 g of salt per 50 grams of sea water what is its

iragen [17]

Answer:

1 cm3 is = 1 ml. Therefore 1000 g of seawater = 973.71 mL.

Explanation:

Seawater salinity will vary from place to place and with the temperature of the seawater. Of course the composition of dissolved substances in seawater, along with salt that is, will also vary from place to place.

On average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand. This means that for every 1 litre (1000 mL) of seawater, there are 35 grams of salts (mostly, but not entirely, sodium chloride) dissolved in it.

Seawater has an average density of 1.027 g/cm3, but this varies with temperature and salinity over a range of about 1.020 to 1.029.

5 0

4 years ago

A gas in a cylinder is held at a constant pressure of 1.80 * 105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3. The in

Lilit [14]

Answer:

The work done is $W=-0.9*10^5J$ and it is negative because the exterior is doing work on the gas.

The value of the heat flow is $Q=2.3*10^5J$ and is directed out of the gas, as this is being cooled and compressed.

Explanation:

The work done by the gas can be written as

$W=\int\limits^ {V_{c}}_{V_{h}}{P}\, dV$

and all the data is given in the problem, so

$W=P(V_{c}-V_{h})=-0.9*10^5J$

is the work necessary to compress the gas in the cylinder, wich is negative because the exterior is doing work on the gas.

Now, from first law of thermodynamics, we have that

$\Delta U=Q+W$

<em>where $\Delta U$ is the internal energy difference, $W$ is the calculated work, and $Q$ is the heat flow that we wish to know</em>, then

$Q=\Delta U-W=1.4*10^5J+0.9*10^5J=2.3*10^5J$

And the direction of the heat flow is outside of the gas, as it is being cooled and compressed.

Finally, it doesn't matter whether the gas is ideal or not, because we never use the Ideal Gas hypotesis, that $PV=nRT$ , and $P$ is constant.

6 0

3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

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Kitty [74]

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8 0

3 years ago

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