A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.
<u> The image produced by the lens is (b) inverted and real</u>
Explanation:
A real image occurs where the rays converge.
Real images can be produced both by the concave mirrors or converging lenses, but the condition is that the object of consideration is always placed far away from the mirror or the lens than the focal point, and thus the real image produced is inverted.
A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.
<u> The image produced by the lens is (b) inverted and real</u>
Answer:
voltage is 1.38 V
Explanation:
given data
weights = 150 gm = 0.150 kg
sensitivity = 0.5 volts/Newton
to find out
large a voltage
solution
we know here in by symmetry so here in x axis
F3 will be
F3 = 2mgcos(20)
F3 = 2(0.150)(9.8)cos20
F3 = 2.76 N
so volatge is
voltage = 2.76×sensitivity
voltage = 2.76×0.5
voltage = 1.38
so voltage is 1.38 V
Answer:
huh,? can you explain the question more please
We have:
s=3,7km (3700m)
u=0
v=26m/s
a=
t=
We look up for a formula to solve for a:
v²=u²+2as u²=0 so
v²=2as
(26m/s)²=2a3700m
676/7400=a
a=0,09m/s²
