Answer:
None, if air resistance is ignored.
Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity..
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.
Answer:
The distance is 55.636 billion miles, or 528.2 AU.
Explanation:
Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

or 55.636 billion miles.
Since 1 astronomical unit (AU) is 93 million miles, that distance is also:

Sound waves travel faster through <em>solids</em> than they do through gases or liquids. <em>(C) </em>They don't travel through vacuum at all.
Example:
Speed of sound in normal air . . . around 340 m/s
Speed of sound in water . . . around 1,480 m/s
Speed of sound in iron . . . around 5,120 m/s
None of the choices is a force. 'a' and 'b' are speeds. 'C' and 'd' are accelerations. ... The steady force of gravity is 9.8 newtons PER KILOGRAM of mass. ... The question is written by someonewho very much wants to discourage anyone interested in Physics.
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;

The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;