Answer:
i). Inverted
ii). Magnification of the image = -0.5
iii). Real
Explanation:
As shown in the ray diagram attached,
An object AB has been placed in front of converging lens (convex lens) at u = 30 cm.
F (Focus) of the lens is at 10 cm. So F = 10 cm
By analyzing the ray diagram we can measure the distance of the image on the other side of the lens (By counting the small blocks of the graph)
V = 15 cm
Characteristics of the image is:
i) Inverted
ii) Magnification of the image = 
= -0.5
ii) Real
The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Given the data in the question;
Since the brick was initially at rest before it was dropped,
- Initial Velocity;

- Height from which it has dropped;

- Gravitational field strength;

Final speed of brick as it hits the ground; 
<h3>Velocity</h3>
velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.
To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Learn more about equations of motion: brainly.com/question/18486505
C.
Newton’s Second Law is F=ma (force is equal to the mass multiplied by acceleration), however, the equation can be rearranged to isolate and calculate mass from force over acceleration. Therefore, m=F/a
Answer:
A
Explanation:
Straight line with a negative slope
On a velocity_time graph
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]


=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v


charge on 2.0μf capacitor is


=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
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