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
Answer:
copper wire, aluminum can r not the answers
Answer:
A) E = 0N/C
B) 0i + 0^^j
C) F = 0N
D) 0^i + 0^j
Explanation:
You assume that the rings are in the zy plane but in different positions.
Furthermore, you can consider that the origin of coordinates is at the midway between the rings.
A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.
You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:
(1)
k: Coulomb's constant = 8.98*10^9Nm^2/C
Q: charge of the ring
r: perpendicular distance to the center of the ring
R: radius of the ring
You use the equation (1) to calculate the net electric field at the midpoint between the rings:

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.
B) The direction of the electric field is 0^i + 0^j
C) The magnitude of the force on a proton at the midpoint between the rings is:

D) The direction of the force is 0^i + 0^j
Choice-'b' says the formula for kinetic energy in words.
KE = (1/2) · (M) · (S²)
Answer:
I think it's d) down the page