Answer:
The answer is can speed an object direction!!!!
Answer:
(a) 0.17 m
(b) 5.003 m
(c) 6.38 ×
N
(d) 7.37 ×
N
Explanation:
(a) The minimum value of
will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.
(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

<em>Hence, the maximum distance is 5.002 m</em>
(c) For minimum magnitude we use the minimum distance calculated in (a)
Minimum Distance = 0.17 m
For electrostatic force= 

×
(d) For maximum magnitude, we use the maximum distance calculated in (b)
Maximum Distance = 5.002 m
Using the formula for electrostatic force again:
F = 
F= 7.37×
N
Answer:
true
Explanation:
a wheelbarrow has its load situated between the fulcrum and the force the wheel Barrow is 2nd class because of its resistance between the force and the axis
15 miles to kilometers would be: 24.14 kilometers
Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;

Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J