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2 years ago

A new moon is discovered orbiting Neptune with an orbital speed of 9.3 103 m/s. Neptune's mass is 1.0 1026 kg. What is the

Physics

1 answer:

grandymaker [24]2 years ago

3 0

Answer:

The radius of the moon's orbit, R = 7.715 x 10⁷ m

The orbital period of the moon, T = 14.48 hr

Explanation:

Given,

The orbital speed of the moon, v = 9.3 x 10³ m/s

The mass of Neptune, M = 1.0 x 10²⁶ Kg

The orbital velocity of the moon is given by the relation

$v = \sqrt{\frac{GM}{r+h}}$ m/s

Where,

r + h = R → radius of the moon's orbit

Therefore squaring the above equation and solving for r +h. The equation becomes

r + h = GM / v²

Substituting the given values in the above equation

R = (6.673 x 10⁻¹¹ x 1.0 x 10²⁶) / (9.3 x 10³)²

= 7.715 x 10⁷ m

The radius of the moon's orbit, R = 7.715 x 10⁷ m

The orbital period of the moon is given by the relation

T = 2π/ω

= 2πR/v ∵ ω = v/r

Substituting the values in the above equation

T = (2 x π x 7.715 x 10⁷ )/ 9.3 x 10³

= 52125.72969 s

= 14.48 hr

The orbital period of the moon, T = 14.48 hr

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For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
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2 years ago

What is the energy of the electron in a helium ion with a charge of +1 in an orbit with a value of n=4?

goldenfox [79]

Answer:

The energy of the electron in a helium ion is $-5.44\times10^{-19}\ J$

Explanation:

Given that,

Charge = +1

Value of n = 4

Suppose $k=2.179\times10^{-18}\ J$

We need to calculate the energy of the electron in a helium ion

Using formula of energy

$E_{n}=-k\dfrac{z^2}{n^2}$

Put the value into the formula

$E_{n}=-2.179\times10^{-18}\times\dfrac{2^2}{4^2}$

$E_{n}=-5.44\times10^{-19}\ J$

Hence, The energy of the electron in a helium ion is $-5.44\times10^{-19}\ J$

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2 years ago

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ

Dovator [93]

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$6.00^2-c^2=\dfrac{w^2}{(20.1)^2}$

$36-c^2=\dfrac{w^2}{404.01}$ ....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$9.00^2-c^2=\dfrac{w^2}{(11.2)^2}$

$81-c^2=\dfrac{w^2}{125.44}$ ....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

$45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}$

$45=w^2(0.00549)$

$w^2=\dfrac{45}{0.00549}$

$w=\sqrt{\dfrac{45}{0.00549}}$

$w=90.5$

We need to calculate the current speed

Using equation (I)

$36-c^2=\dfrac{(90.5)^2}{(20.1)^2}$

$36-c^2=20.27$

$c^2=20.27-36$

$c=\sqrt{15.73}$

$c=3.96\ m/s$

(b). We need to calculate the shortest time

Using formula of time

$t=\dfrac{d}{v}$

$t=\dfrac{90.5}{6}$

$t=15.0\ sec$

We need to calculate the distance

Using formula of distance

$d=vt$

$d=3.96\times15.0$

$d=59.4\ m$

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

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2 years ago

A ball is thrown into the air and follows the parabolic path indicated by the dashed line. Use the arrows provided to indicate t

avanturin [10]

Answer:

Trajectory is the path of a projectile in air. Any tangential to the curve shows it velocity and the slope of the tangential line will be it acceleration.

Explanation:

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2 years ago

A planet orbits a star along an elliptical path from point X to point Y, as shown in the figure. In which of the following syste

8090 [49]

Answer:

The correct answer is Option D, the closed system containing the planet and the star.

Explanation:

To start, we need to define mechanical energy: the energy an object has from its motion and position.

The fundamental principle in physics is that the total energy in a closed system stays constant, even if it transforms. By saying "closed system," we refer to a system isolated from its surroundings. Energy never leaves the system; it only moves from one part to another.

This statement only applies to closed systems, however. An open system that interacts with its environment works differently. Energy may enter and leave the system through interaction with external forces, and this includes mechanical energy. For this reason, Option A and Option B are incorrect.

The remaining two options, C and D, only vary with the objects in the closed system. Option D includes the star; Option C does not.

However, we should take a closer look at Option C. Can an object have potential energy with itself? No, it cannot. It only has potential energy with other bodies. If the system is defined as the planet only, the only type of energy present is kinetic energy. We know a planet orbiting a star has more kinetic energy near and more gravitational potential energy further from its star. Thus it has less kinetic energy further from its star and less mechanical energy. Because of this, Option C is incorrect.

The only answer left is Option D. If we define the planet and star as a closed system, we find no net external force acting on it. Consequently, it obeys the law of conservation of energy. From prior reasoning, we know mechanical energy includes potential energy and kinetic energy and that the amounts of these energies vary with its orbit. As a result, mechanical energy is always conserved and always the same. In the end, the correct answer is Option D.

4 0

2 years ago