When running a marathon what kind of energy transformation takes place

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

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

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

or

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago

Supposing d(t) is known to have value D,

creativ13 [48]

Answer:

The procedure is: solve the quadratic equation for $t$ .

Explanation:

This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:

$d(t)=d_0+v_0t+at^2/2$

That is a quadratic equation, where the independent variable is the time $t$ .

Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for $t$ .

To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:

$(a/2)t^2+v_0t+(d_0-D)=0$

Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:

$t=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

Where:

$a=-a/2\\ \\ b=v_0\\ \\ c=d_0-D$

That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.

5 0

3 years ago

A physical change can be distinguished from a chemical change because during a physical change- A. the properties of the substan

joja [24]

Answer:

a

Explanation:

yw

3 0

2 years ago

. A non-uniform positive line charge of length 2 m is put along the x-axis as shown in the figure, where x0=1.5 m. The linear

PilotLPTM [1.2K]

Answer:

$E = 7.2*10^{10}N/C$

Explanation:

The differential electric field $dE$ due to differential charge $dQ$ at distance $x$ from the origin is

$dE = k\dfrac{dQ}{x^2}$

but since $dQ = \lambda dx = 4x^2dx$ we have

$dE = k\dfrac{4x^2dx}{x^2}$

$dE = 4k\: dx$

integrating this from $x_0$ to $x_0+L$ we get

$E = \int^{x_0+L}_{x_0} {4k} \, dx$

$E = 4k[(x_0+L)-x_0]$

$E =4kL$

putting in $k = 9*10^9Nm^2/C^2$ and $L =2m$ we get

$\boxed{E = 7.2*10^{10}N/C.}$

5 0

3 years ago

A push or pull is called a force. Unbalanced forces can cause objects to move in many ways but not to move faster. B) move slowe

Allushta [10]

Answer:

but not to

D). float in the air

6 0

3 years ago