Answer:
cross out the false piece in blue and write the true piece in red
<u>We are given:</u>
Mass of Neptune = 1.03 * 10²⁶ kg
Distance from the center of Neptune (r) = 2.27 * 10⁷
now, computing the value of the acceleration due to gravity (g)
<u>Finding g:</u>
We know the formula:
g = G(mass of planet) / (r)²
g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]
g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)
which can be rewritten as:
g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15
g = (6.87 * 10¹⁵⁻¹⁴) / 5.15
g = (6.87/5.15) * 10
g = 1.34 * 10
g = 13.4 m/s² <em>(approx)</em>
Answer:
λ = 5.65m
Explanation:
The Path Difference Condition is given as:
δ=
;
where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.
m = no of openings which is 2
∴δ= 
n is the index of refraction of the medium in which the wave is traveling
To find δ we have;
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 
δ= 82.15 -73.68
δ= 8.47
Again remember; to calculate the wavelength of the ocean waves; we have:
δ= 
δ= 8.47
8.47 = 
λ = 
λ = 5.65m
Answer:42.43m/s
Explanation:According to vf=vi+at, we can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s
A contact force is a type of force which act on an object by coming in contact with the object. Examples of contact force that acts through a force field are: applied force, frictional force, air resistance force, tension, spring force, etc.
Examples of forces that act through a force field are gravitational force, electromagnetic force, the weak interaction and the strong interaction.