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Alex Ar [27]
3 years ago
9

Moons are natural satellites. They do not orbit the Sun directly. Instead, they move in an orbit around a celestial body, such a

s a planet. Which celestial bodies have one or more moons?
asteroids
comets
dwarf planets
meteoroids
planets

( You can pick more then one answer)
Physics
2 answers:
ankoles [38]3 years ago
8 0

Answer: The correct answer is- Planet and dwarf planet.

Explanation -

The natural satellite that moves around a celestial body is known as moon. The celestial body is specific to planets and dwarf planets.

For instance, planet Earth has one moon, Mars has two moons, and dwarf planet (pluto) has more than two moons.

Thus, planets and dwarf planets are the right answer.


iogann1982 [59]3 years ago
4 0
Planets and Dwarf Planets
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Hey, guys I kinda need help with this question, 1. The number of atoms of each element are not equal on both sides of the equati
Xelga [282]

Answer:

The same number of atoms of each element must appear on both sides of a chemical equation. However, simply writing down the chemical formulas of reactants and products does not always result in equal numbers of atoms. You have to balance the equation to make the number of atoms equal on each side of an equation.

Explanation:

I hope thats what u needed.

5 0
3 years ago
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11. Consider the velocity vs time graph below. Which object is not moving at time = 0?
MAXImum [283]
I’m guessing it’s all of the above
8 0
3 years ago
A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300
raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

<em>A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:</em>

<em>a)</em><em> Find the work done by the force.</em>

<em>b)</em><em> Find the speed of the mass at the end of the 3.5 meters.</em>

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force (W_{F}), measured in joules, is:

W_{F} = F\cdot \Delta s \cdot \cos \theta (1)

Where:

F - External constant force exerted on the mass, measured in newtons.

\Delta s - Horizontal travelled distance, measured in meters.

\theta - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that F = 45\,N, \Delta s = 3.5\,m and \theta = 30^{\circ}, then the work done by the force is:

W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}

W_{F} = 136.400\,J

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2}) (2)

Where:

m - Mass, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the mass, measured in meters per second.

If we know that W_{F} = 136.400\,J, m = 2\,kg and v_{1} = 0\,\frac{m}{s}, then the final speed of the mass is:

\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}

v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}

v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }

v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }

v_{2} \approx 11.679\,\frac{m}{s}

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

3 0
3 years ago
Frequency, period and wavelength<br> 11th grade high school physics
zzz [600]

Answer:0.38

Explanation:

the formula is f = c / λ

so f= 2.5/6.5

and that equals 0.38 46 and so on so i just rounded it

6 0
3 years ago
A 750 watt hairdryer is used for 60 seconds. how many joules of energy are used?
daser333 [38]
We know, Power = Energy / time
Here, P = 750 Watt
t = 60 sec

Substitute their values, 
750 = E / 60
E = 750 * 60
E = 45,000 J

In short, Your Answer would be 45,000 Joules

Hope this helps!
4 0
4 years ago
Read 2 more answers
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