Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ = = = 3.61cm = 0.036m
r₂ = = = 5cm = 0.05m
electric potential V =
change in potential ΔV =
ΔV = , where 2.00μC
ΔV =
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ ×
ΔV= 2.789×10⁵
= ΔV × q₃
ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s
Answer:
solution:
when the engine are fired the rocket has a linear constant acceleration motion:
is the final velocity of the rocket
when the engines are fired it become equal to the initial velocity of the rocket,
<em>when the engines are shut off</em>
<em></em><em></em>
solve eq(1) and eq(2) we find
solving for =364.65 m
Where is the distance travelled by the rockets for shutting off the engine
<em>when the engines are fired:</em>
<em></em><em></em>
NOTE:
DIAGRAM IS ATTACHED
<span>It all starts from a radio with a loudspeaker . It produces and converts electrical energy and make this a formed of kinetic energy coming out of the speaker. Then, when this kinetic energy is moving and transferred to surroundings or air, it will let people feel the motion being created by the air. Finally, it reaches the people and hears this as sound.
This is an example on how we trace the transformation of on signal from radio and how it reaches to people as sound or as a music.</span>