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Irina18 [472]
3 years ago
15

Use the following formula to answer the following statement. Round to the nearest tenth.

Physics
1 answer:
just olya [345]3 years ago
3 0
I need help my self lol XD
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The Trojan asteroids are found:_________ a) orbiting around the Kuiper Belt body Hector. b) beyond Neptune, with orbits similar
xenn [34]

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.

5 0
3 years ago
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Answer this plsss<br> A mobile phone operates at 900 MHz. <br> What wavelength does it use?
Aleks [24]

Answer:

The wavelength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters.

7 0
3 years ago
Compare the mass on Neptune (68.11) to the weight on Neptune (905.863). What is the difference? Why is there a difference?
Alchen [17]
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7 0
3 years ago
Write down applications of mechanics​
harkovskaia [24]

Answer:

Explanation:

applied Mechanics and its Growing Utilisation of Theoretical Mechanics.\

Structural Engineering.

Hydraulics.  

Mechanical Engineering.  

External Fluid Dynamics.

Planetary Sciences.

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7 0
3 years ago
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
Pepsi [2]

Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
4 years ago
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