Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Answer:
<h3>
30.66m</h3>
Explanation:
Using the equation of motion formula
where;
S is the height to which the ball rises
u is the initial velocity of the ball = 0m/s
a is the acceleration due to gravity = 9.81m/s²
t is the time taken by the ball in air = 5.0s
Note that the time to rise to the peak is one-half the total hang-time = 5.0/2 = 2.5s
Substituting the given parameters into the formula above to get S:

This means that the ball rises 30.66m before it reaches its peak.
First of all, you didn't tell us WHO measured the "10 years".
If it was the people on Earth, then 10 years passed according to them.
If it was 10 years on the space traveler's clock, then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.
If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by
√(1 - v²/c²) .
You said that v/c = 0.85 .
v²/c² = (0.85)² = 0.7225
1 - v²/c² = 1 - 0.7225 = 0.2775
√(1 - v²/c²) = √0.2775 = 0.5268
If one clock counts up 10 years, then the other one counts up
(10years) / 0.5268 = <em>18.983 years </em>
I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.