The two will fall at the same speed and reach the surface at the same time. This is because the two will experience the same gravitational acceleration on the moon. However, on the earth surface the two will land on the surface of the earth at the same time due to air resistance such that the egg will experience a higher air resistance than the hammer. On, the moon, where there is no noticeable atmosphere there is no air resistance on either object and both fall at the same speed. It is also important to note that their mass doesn't affect their speed.
Answer:
a moving object will keep moving if not stopped
the sun being at the center of the solar system
Explanation:
Galileo is known for being the first person make a telescope, there fore being the first person to see that the sun is in the center of the solar system. he also came up with the theory that if something is pushed, it would keep moving until stopped by another force. For example, say you drop your pencil, it keeps falling until it hits the ground. That is exactly what Galileo did in his Leaning Tower of Pisa experiment and found that theory to be true.
Answer: 
Explanation:
This problem can be solved using the Third Kepler’s Law of Planetary motion:
<em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
<em />
This law states a relation between the orbital period 
of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size 
of its orbit:
(1)
Where:
is the period of the orbit of the exoplanet (considering 
)
is the Gravitational Constant and its value is 
is the mass of the star
is orbital radius of the orbit the exoplanet describes around its star.
Now, if we want to find the radius, we have to rewrite (1) as:
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%5E%7B2%7D%7D%7D)
(2)
![a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%28122044320s%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%283.59%2810%29%5E%7B30%7Dkg%29%7D%7B4%5Cpi%5E%7B2%7D%7D%7D)
(3)
Finally:
This is the radius of the exoplanet's orbit
KE = 1/2 * m* v^2
= 1/2 * 40 * 225
= 4500 J
hope it helped
Answer:
4.92°
Explanation:
The banking angle θ = tan⁻¹(v²/rg) where v = designated speed of ramp = 30 mph = 30 × 1609 m/3600 s = 13.41 m/s, r = radius of curve = 700 ft = 700 × 0.3048 m = 213.36 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the variables into the equation, we have
θ = tan⁻¹(v²/rg)
= tan⁻¹((13.41 m/s)²/[213.36 m × 9.8 m/s²])
= tan⁻¹((179.8281 m²/s)²/[2090.928 m²/s²])
= tan⁻¹(0.086)
= 4.92°
