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Fynjy0 [20]
3 years ago
11

The Trojan asteroids are found:_________ a) orbiting around the Kuiper Belt body Hector. b) beyond Neptune, with orbits similar

to Pluto's. c) sixty degrees ahead or behind Jupiter, sharing its orbit about the Sun. d) with the others, between Mars and Jupiter; their red color gives them their name.
Physics
2 answers:
xenn [34]3 years ago
5 0

Answer:

The correct option is;

c) sixty degrees ahead or behind Jupiter, sharing its orbit about the Sun.

Explanation:

The Trojan asteroids are the Jupiter trojans consists of asteroid that are on the same orbit as Jupiter while moving around the Sun. The Trojans can be located at the points Lagrange points L4 and L5, which are 60° ahead and 60°  behind Jupiter's orbit respectively.

The first Trojan asteroid to be detected was 588 Achilles by Max Wolf in 1906. At at October, the total number of the identified Trojan asteroid was  7,040.

Katarina [22]3 years ago
4 0

Answer:

c) sixty degrees ahead or behind Jupiter, sharing its orbit about the Sun.

Explanation:

Trojan Asteroids are one type of co-orbital object. It is a small body that shares the orbit of a larger one, they remain in a stable orbit at approximately 60° ahead or behind the main body near one of its lagrangian points L4 and L5, they can share the orbits of planets or of large moons. Due to the perturbations by other planets, their distributions are elongated along the orbit, there are currently over 4,800 known trojan asteroids associated with Jupiter.

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Which of the following does NOT describe the conversion of potential to kinetic energy? Question 12 options:
Marianna [84]

The option “The sum emitting light energy” does not describe the conversion of potential energy to kinetic energy.

Answer: Option A

<u>Explanation: </u>

The potential energy, the object's energy at rest while the kinetic energy is the energy the object in motion. So among the options given, a rocket leaving the Earth's atmosphere consists of conversion of potential energy as the thrusting force to kinetic energy of the movement of the rocket to space.

Similarly the stretching of rubber band is a form of elastic potential energy and when it is released, the band will vibrate which will be the converted kinetic energy.

Also, the avalanche is the falling of ice berg which were once in rest, so the potential energy of the ice berg will get converted to kinetic energy during avalanche.

The option that is the sun emitting light does not convert potential to kinetic energy, instead it converts potential to light energy.

6 0
3 years ago
Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
You can also enter units that are combinations of other units. keep in mind that you have to indicate the multiplication of unit
omeli [17]
Watch this !  You've never seen anything like it !
No wait.  I'm pretty sure you have.


Weight = (mass) · (gravity)

Weight = (10 kg) · (9.8 m/s²)

Weight = (10 · 9.8) kg·m/s²

Weight = 98 Newtons
7 0
4 years ago
How much GPE is stored when an 80kg astronaut climbs to the top of a 5m high lunar lander? The gravity strength on the moon is 1
8_murik_8 [283]

Answer:

The GPE, stored is 640 Joules

Explanation:

The given parameters are;

The given mass of the astronaut, m = 80 kg

The height of the top of the lunar lander to which the astronaut climbs, h = 5 m

The gravity strength on the moon, g = 1.6 N/kg

The Gravitational Potential Energy, GPE, stored is given according to the following equation;

GPE stored = m·g·h

Therefore, by substituting the known values, we have;

GPE Stored = 80 kg × 1.6 N/kg × 5 m = 640 Joules

The GPE, stored = 640 Joules.

6 0
3 years ago
The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current
Komok [63]

Answer:

I=0.047A

Explanation:

Let's use Ohm's law:

V=IR  

or

I=\frac{V}{R}   (1)

Where:

V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

R=\rho*\frac{l}{A}    (2)

Where:

R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}

Keep in mind that the electrical resistivity of the gold is a known constant which is \rho_g_o_l_d=2.35*10^{-8} and the cross sectional area of the conductor is calculated as:

A=\pi *(r^{2})=\pi  *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} }  =14.96056465\Omega

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

I=\frac{0.7}{14.96056465}\approx0.047A

7 0
3 years ago
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