Enlarged real image.
A converging lens always has a positive focal point so the light rays always converge on the real side of the lens, creating a real image.
It will also be enlarged because when you draw a ray diagram you can see that the image will be about two times or so larger depending on where the position of the object actually is. Usually when an image is reduced, it is when it is a diverging / concave lens.
<span>To determine the magnitude and the direction of the resultant force, we assume that the forces are in XY coordinate plane and the angles that are given are from the x axis.
</span>
<span>The 110 N force is said to act at 90 deg which means it is along the Y axis. The </span><span>55 N force is said to act at 0 deg which means it is along the X axis. so, a right angle is made by the two forces. Thus, the
</span>
<span>X component of the resultant force = 55 N </span>
<span>Y component of the resultant force = 110 N </span>
<span>Magnitude of the resultant force would be calculated as follows:
R = √(Fx^2 + Fy^2)
R = √(55^2 + 110^2) </span>
<span>R = √(15125) </span>
<span>R = 123 N </span>
<span>The resultant force would have its terminal side in the x-axis. We calculate angle θ as follows:
</span>
<span>tan θ = Fy/Fx </span>
<span>tan θ = 110 N /55 N = 2
</span>θ = arctan(2)
θ <span>= 63.4 degrees
</span>Therefore, the m<span>agnitude of the resultant force is 123 N and the direction would be at an angle of 63.4 degrees.</span>
Answer:
B) The same as the momentum change of the heavier fragment.
Explanation:
Since the initial momentum of the system is zero, we have
0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.
0 = p + p'
p = -p'
Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0 = p
The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'
Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'
<u>So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment. </u>
So, option B is the answer
Answer:
Option B. 3.0×10¯¹¹ F.
Explanation:
The following data were obtained from the question:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:
Capacitance (C) = Charge (Q) / Potential difference (V)
C = Q/V
With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:
Potential difference (V) = 100 V.
Charge (Q) = 3.0×10¯⁹ C.
Capacitance (C) =..?
C = Q/V
C = 3.0×10¯⁹ / 100
C = 3.0×10¯¹¹ F.
Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.
