Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer
Answer
given,
time interval = 11.3 s
a) initial velocity, vi = 15 m/s
final velocity, v_f = -5.30 m/s
a = -1.79 m/s²
the direction is along left side
b) initial velocity, vi = -5.30 m/s
final velocity, v_f = -15 m/s
a = -0.858 m/s²
the direction is along left side
c) initial velocity, vi = 15 m/s
final velocity, v_f = -15 m/s
a = -2.65 m/s²
the direction is along left side
Answer: the answer is D
Explanation:
When we look at an object and see its color, we are seeing all of the light that reflects off of that object. Red objects reflect red light, green objects reflect green light, and so on. But what happens to the rest of the colors that hit that object? They get absorbed!
Projectile motion is characterized by an arc-shaped direction of motion. It is acted upon by two vector forces: horizontal component and vertical component. The horizontal component is in constant velocity motion, while the vertical component is in constant acceleration motion. These two motions are independent of each other.
Now, the total velocity of the space probe at the end of the projectile motion is determined through this equation:
V = √(Vx² + Vy²)
where Vx is the velocity in the horizontal direction and Vy is the velocity in the vertical direction.
Let's find Vx first. Assuming that the space probe was launched at an angle horizontal the Earth's surface, the launching angle is 0°. Thus, the initial velocity is 2.44×10⁴ m/s.
For Vy, the free falling motion is
Vy = √(2gh), where g is 9.81 m/s² and h is the distance traveled vertically by the space probe which represents the radius of the Earth equal to 6.37×10⁶ meters. Therefore,
V = √{(2.44×10⁴)² + [√(2×9.81×6.37×10⁶ )]²}
V = 26,839.14 m/s
