Answer:
0.28 m
Explanation:
The following data were obtained from the question:
Force (F) = 5×10¯⁶ N
Charge 1 (q₁) = 6.7×10¯⁹ C
Charge 2 (q₂) = 6.7×10¯⁹ C
Electrical constant (K) = 9×10⁹ Nm²C¯²
Distance apart (r) =?
Thus, the distance between the two charges can be obtained as follow:
F = Kq₁q₂/r²
5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²
5×10¯⁶ = 4.0401×10¯⁷ / r²
Cross multiply
5×10¯⁶ × r² = 4.0401×10¯⁷
Divide both side by 5×10¯⁶
r² = 4.0401×10¯⁷ / 5×10¯⁶
Take the square root of both side
r = √(4.0401×10¯⁷ / 5×10¯⁶)
r = 0.28 m
Therefore, the distance between the two charges is 0.28 m
Answer:
- The procedure is: solve the quadratic equation for
.
Explanation:
This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:
That is a quadratic equation, where the independent variable is the time
.
Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for
.
To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:
Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:
Where:

That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.
Explanation:
Work is the product of force and distance.
W = F×d
W = (22 N) × (16 m)
W = 352 J
Considering that while traveling on a road with a<u> final speed of 15 m/s</u>, and an<u> initial speed of 24 m/s</u>, with a given time <u>of 12 seconds.</u>
To calculate the acceleration, we apply the following formula:
α = Vf - Vo/t
We add our data into the formula and solve:
α = 15 m/s - 24 m/s/12 sec
α = -0.75 m/s²
Therefore, the acceleration of the car is -0.75 m/s².
<h2>Skandar</h2>
Frequency =1/period
Freq= 1/6= 0.17 Hertz