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2 years ago

The law of suggests that the orbit of planets is not circular but.

2 answers:

7 0

Newton's law of universal gravitation ... massaged and manipulated
with a lot of calculus and geometry ... predicts that closed gravitational
orbits are ellipses.

The circle is such a precise special kind of ellipse that a circular orbit
never occurs. The chance of it is like the chance of flipping a coin and
having it land standing up on its edge ... possible, but very improbable.

anygoal [31]2 years ago

5 0

Answer:

Kepler's 2nd law describe the shape of orbit

Explanation:

Kepler's 2nd law describe the shape of orbit

A planet's orbit's around which it revolve around the Sun is not a circle. It's an ellipse — a ring that's "flattened." The Sun (which is also called the planet's center) holds one ellipse target. A concentration is among two inner points helping to decide an ellipse's form.

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Which of the following is most likely to happen when a plate carrying oceanic crust collides with a plate carrying continental c

gavmur [86]

ASAP here you go It's A

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2 years ago

What is the speed of sound in solid? <br>

sasho [114]

The speed of sound in a solid would be 6000 metres per second

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2 years ago

Read 2 more answers

A positively charged rod is brought close to one end of an uncharged metal rod but does not actually touch it. What type of char

svlad2 [7]

Answer:

The end of the neutral rod which is the closest part to the charged rod would acquire a negative charge.

Explanation:

One of the rods is positively charged and one of them is neutral.

And the important part is that <u>they do not touch one another</u>, but get close to each other.

In this case, the end of the neutral rod which is the closest part to the charged rod would acquire a negative charge. This is because of the Coulomb's Law. The opposite charges exert an attractive force to each other. The positive charges attract the negative charges on the neutral rod.

7 0

3 years ago

Since sound is a mechanical wave it needs a ...... to travel through

Leni [432]

Answer:

medium

Examples:

A medium is a thing that the sound wave would travel through, like sound waves traveling through air. A mechanical wave cannot travel without a medium, like if the wave were trying to travel through a vacuum.

6 0

2 years ago

A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference

Andre45 [30]

a) $400 \Omega$

b) 0.43 V

c) 0.44 %

Explanation:

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

$V=E-Ir$ (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when $R_1=550 \Omega$ , $V_1=0.25 V$

Using Ohm's Law, the current is:

$I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A$

2) when $R_2=1000 \Omega$ , $V_2=0.31 V$

Using Ohm's Law, the current is:

$I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A$

Now we can rewrite eq.(1) in two forms:

$V_1 = E-I_1 r$

$V_2=E-I_2 r$

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

$V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega$

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

$V=E-Ir$

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

$V=0.25 V$ is the voltage

$I=4.5\cdot 10^{-4}A$ is the current

$r=400\Omega$ is the internal resistance

Solving for E,

$E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V$

In this part, we are told that the area of the cell is

$A=4.0 cm^2$

While the intensity of incoming radiation (the energy received per unit area) is

$Int.=5.5 mW/cm^2$

This means that the power of the incoming radiation is:

$P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W$

This is the power in input to the resistor.

The power in output to the resistor can be found by using

$P'=I^2R$

where:

$R=1000 \Omega$ is the resistance of the resistor

$I=3.1\cdot 10^{-4} A$ is the current on the resistor (found in part A)

Susbtituting,

$P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W$

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

$\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%$

7 0

2 years ago