Answer:
<u>Principal</u><u> </u><u>focus</u><u> </u><u>of</u><u> </u><u>concav</u><u>e</u><u> </u><u>lens</u><u> </u><u>-</u><u> </u>
★ The point at which rays parallel to principal axis coming from infinity appear to converge after being refracted from concave lens is called the principal focus of concave lens.
<em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em>
• <u>Additional</u><u> information</u><u> </u><u>-</u><u> </u>
★ Principal focus - A number of rays parallel to the principal axis after reflection from a concave mirror meet at a point on the principal axis or appear to come from a point after reflection from a convex mirror on the principal axis. This is called principal focus.
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:
15
Explanation:
P=W/T
T=6sec
W=?
F=60N
S=18m
W=F X S. .s indicate displacement
W=60x18
W=108
So p=108 j/6sec
P=15watt
Answer:
a) Not Accurate
b) Not Accurate
c) Accurate
d) Accurate
Explanation:
Part a
Not Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m
Part b
Not Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m
Part c
Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m by varying the phase difference between two waves she can achieve the desired results.
Part d
Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m by varying the phase difference between two waves she can achieve the desired results.