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3 years ago

How do earths movements affect our view of the star

2 answers:

7 0

In the solar system the Earth rotates around the sun and the moon rotates around the Earth. this cause half of the earth to be bright at a time.. the moon has different phases which causes the earth to be affected

Rzqust [24]3 years ago

4 0

As the Earth makes a full revolution, we see the sun and no stars, except the sun, and when our section of Earth is facing away from the sun we can see other starts.
Also, as the Earth revolves around the sun, certain stars can be seen and certain stars go out of view.

Hope this helps!!(:

You might be interested in

Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de

Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 $\frac{m}{s}$

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa. Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido, en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.

v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo $A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4}$ , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

$\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2$

En este caso:

D1: 0.06 m
v1: ?
D2: 0.04 m
v2: 2.6 m/s

Reemplazando:

$\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}$

Resolviendo:

$v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}$

$v1=\frac{(0.04m)^{2} }{(0.06m)^{2} }*2.6\frac{m}{s}$

v1= 1.156 $\frac{m}{s}$

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 $\frac{m}{s}$

8 0

3 years ago

When a 360 nF air capacitor is connected to a power supply, the energy stored in the capacitor is 1.85 x 10-5 J. While the capac

GaryK [48]

Answer:

(a) Approximately $10.1\; {\rm V}$ .

Explanation:

Let $C$ denote the capacitance of a capacitor. Let $V$ be the potential difference (voltage) between the two plates of this capacitor. The energy $E$ stored in this capacitor would be:

$\displaystyle E = \frac{1}{2}\, C\, (V^{2})$ .

Rearrange this equation to find an expression for the potential difference $V$ in terms of capacitance $C$ and energy $E$ :

$\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}$ .

$\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}$

The capacitance $C$ of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):

$\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}$ .

Given that the energy stored in this capacitor is $E = 1.85 \times 10^{-5}\; {\rm J}$ , the potential difference across the capacitor plates would be:

$\begin{aligned}V &= \sqrt{\frac{2 \times 1.85 \times 10^{-5}\; {\rm J}}{3.60 \times 10^{-7}\; {\rm F}}} \\ &\approx 10.1\; {\rm V}\end{aligned}$ .

7 0

3 years ago

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

zhannawk [14.2K]

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}$

$v = 16.52 m/s$

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}$

$v = 7.47 m/s$

4 0

3 years ago

Suppose that the resistance between the walls of a biological cell is 6.8 × 109 ω. (a) what is the current when the potential di

GenaCL600 [577]

(a) We can find the current flowing between the walls by using Ohm's law:
$I= \frac{\Delta V}{R}$
where $\Delta V=69 mV=0.069 V$ is the potential difference and $R=6.8\cdot 10^9 \Omega$ is the resistance. Substituting these values, we get
$I=1.01 \cdot 10^{-11} A$

(b) The total charge flowing between the walls is the product between the current and the time interval:
$Q=I \Delta t$
The problem says $\Delta t=0.86 s$ , so the total charge is
$Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C$

The current consists of Na+ ions, each of them having a charge of $e=1.6 \cdot 10^{-19} C$ . To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
$N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions$

4 0

3 years ago

Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the

dangina [55]

Answer:

a. The spheres will attract each other.

Explanation:

When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.

When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.

Hence the two spheres will experience some amount of electrostatic force between them.

4 0

3 years ago