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3 years ago

Describe the motion of the moon in space to produce the different phases of the moon.

Physics

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

Moon slowly moves eastward, rising later each day and passing through its phases:

Explanation:

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How are reflecting telescopes similar to devices that produce laser light?

Novosadov [1.4K]

Answer:a

Explanation:

Because its light

7 0

3 years ago

se lanza un cuerpo desde el origen con velocidad horizontal de 40 m/s, y con un ángulo de 60º. calcular la máxima altura y el al

EastWind [94]

Answer:

1. $h = 244.8 m$

2. $x = 564.8 m$

Explanation:

1. La altura máxima se puede calcular usando la siguiente ecuación:

$v_{f}^{2} = v_{0}^{2} - 2gh$ (1)

Where:

$v_{f_{y}}$ : es la velocidad final = 0 (en la altura máxima)

$v_{0_{y}}$ : es la velocidad inicial horizontal en "y"

g: es la gravedad = 9.81 m/s²

h: es la altura máxima =?

La velocidad incial en "y" se puede calcular de la siguiente manera:

$tan(\theta) = \frac{v_{0_{y}}}{v_{0_{x}}}$

$v_{0_{y}} = tan(60)*40 m/s = 69.3 m/s$

Resolviendo la ecuación (1) para "h" tenemos:

$h = \frac{v_{0_{y}}^{2}}{2g} = \frac{(69.3 m/s)^{2}}{2*9.81 m/s^{2}} = 244.8 m$

2. Para calcular el alcance horizontal podemos usar la ecuación:

$x = v_{x}*t$

Primero debemos encontrar el tiempo cuando la altura es máxima ( $v_{f_{y}}$ = 0).

$v_{f_{y}} = v_{0_{y}} - gt$

$t = \frac{v_{0_{y}}}{g} = \frac{69.3 m/s}{9.81 m/s^{2}} = 7.06 s$

Ahora, como el tiempo de subida es el mismo que el tiempo de bajada, el tiempo máximo es:

$t_{m} = 2*7.06 s = 14.12 s$

Finalmente, el alcance horizontal es:

$x = 40 m/s*14.12 s = 564.8 m$

Espero que te sea de utilidad!

7 0

3 years ago

Mark's uncle is about to have a procedure to treat his prostate cancer. The treatment involves placement of small pellets close

Rus_ich [418]

Answer: brachytherapy

Explanation:

Just answered it right

7 0

3 years ago

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the o

julsineya [31]

Answer:

1) $k=-\frac{1}{46}\approx -0.02$

2) $\textit{Limiting value} = 24$

3) $T(10)\approx \frac{494611944}{6436343}\approx 76.8$

Explanation:

First of all note that

$T_s=24$ is the surroundings temperature, the temperature of the room where the cup of coffee is. Then, the differential equation is:

$\frac{dT}{dt}=k(T-24)$

Also, note that all units are in degrees celsius and minutes. Then, we don't have to convert units. Let's not write units explicitly from now on.

Explanation

1) We have that

$\textit{rate of cooling}=\frac{dT}{dt}=1,\quad T=70$

at some point - the exact time at which this is true doesn't really play any role because the equation doesn't have t on the right hand side. Then, from the equation we get

$1=-(70-24)=46k\Rightarrow k=-\frac{1}{46}\approx -0.02$

The minus comes from considering the temperature must decrease. With this value we can write the equation more explicitely:

$\frac{dT}{dt}=-\frac{1}{46}(T-24)$

2) The coffee is cooling off as time goes by, and it won't get any cooler than 24 degrees celsius because that's the temperature of the room. Then, in the long run, the temperature of the coffee is 24 degrees celsius.

3) Remember that Euler's method consists of using an initial exact measurement to predict what will happen in the future, approximately. There is a formula to make those predictions an it depends on the time step they gave us. Let's compute things first and then I tell you the equations we used.

In this case we know that we start with a 90 degrees celsius cup of coffee, or, in terms of math,

$T(0)=90$

Then, we can predict:

$T(2)\approx 90+2\left[-\frac{1}{46}(90-24)\right]=\frac{2004}{23}\approx 87.1$

Let's use fractions so we don't lose accuracy from now. With this number we can make an approximation of the temperature after 2 more seconds:

$T(4)\approx \frac{2004}{23}+2\left[-\frac{1}{46}\left(\frac{2004}{23}-24\right)\right]=\frac{44640}{529}\approx 84.4$

and then

$T(6)\approx \frac{44640}{529}+2\left[-\frac{1}{46}\left(\frac{44640}{529}-24\right)\right]=\frac{994776}{12167}\approx 81.8$

and then

$T(8)\approx \frac{994776}{12167}+2\left[-\frac{1}{46}\left(\frac{994776}{12167}-24\right)\right]=\frac{22177080}{279841}\approx 79.2$

and finally, the number we wanted to find:

$T(10)\approx \frac{22177080}{279841}+2\left[-\frac{1}{46}\left(\frac{22177080}{279841}-24\right)\right]=\frac{494611944}{6436343}\approx 76.8$

I hope you noticed the pattern to compute the next prediction:

$\textit{next prediction} = \textit{previous one (or exact value if it's the first step)}\\+ h\ast(\textit{right hand side of the differential equation at the previous one})$

5 0

3 years ago

A chimpanzee sitting against his favorite tree gets up and walks 70.9 m due east and 31.9 m due south to reach a termite mound,

DIA [1.3K]

A right triangle is formed by the 70.9 m walked east and 31.9 m walked south.
The legs of this right triangle are 70.9 and 31.9.
The shortest distance between two points is a straaight line. Therefore the hypotenuse of this triangle is going to be that shortest distance.
We can use the Pythagorean Theorem to find the hypotenuse of this triangle.
$a^2+b^2=c^2\\70.9^2+31.9^2=c^2\\5026.81+1017.61=c^2\\6044.42=c^2\\c=\sqrt{6044.42}$

$c=\sqrt{6044.42}\approx\boxed{77.7458681}\ (decimal\ form)\\\\c=\sqrt{6044.42}=\sqrt{\frac{604442}{100}}=\boxed{\frac{\sqrt{604442}}{10}}\ (exact\ form)$

As for the second part of the question, we want to find the angle formed by the hypotenuse and the 31.9 walked east.
We could use any of the three trigonometric ratios here since we know all 3 sides.
sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent
I am going to use tangent, because then I won't have to deal with the hypotenuse and so the answer will be more accurate.

If you haven't already drawn yourself a diagram, now is a good time to.
The side opposite our angle is the 31.9, and the adjacent is 70.9.
Therefore, $\tan(m\angle)=\frac{31.9}{70.9}$ .

We can use inverse trig ratios here to find the measure of our angle.
$\tan^{-1}(\tan(m\angle))=\tan^{-1}(\frac{31.9}{70.9})\\\\m\angle=\tan^{-1}(\frac{31.9}{70.9})\approx\boxed{24.2243851\°\ or\ 0.422795279\ rad}$

4 0

3 years ago