Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;

Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.


f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;

Here, p is the power of the lens.
Put f= 35.71 cm.

p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
Answer:
induced emf = 28.65 mV
Explanation:
given data
diameter = 7.3 cm
magnetic field = 0.61
time period = 0.13 s
to find out
magnitude of the induced emf
solution
we know radius is diameter / 2
radius = 7.3 / 2
radius = 3.65 m
so induced emf is dπ/dt = Adb/dt
induced emf = A × ΔB / Δt
induced emf = πr² × ΔB / Δt
induced emf = π (0..65)² × ( 0.61 - (-0.28)) / 0.13
induced emf = 0.0286538 V
so induced emf = 28.65 mV
Answer:
II) Kitchen waste: Meal leftovers, Banna peelings
Garden Waste: Camote leaves, Kangkong leaves, weeds
Factory: Glass bottles, carton pieces
III) A
IV) Home: Bottles of shampoo, leftover food, syringe
office Gloves
Classroom: containers
Laboratory: empty cartridge
Answer:
D) Since the stars would move from East to West just as the Sun and Moon do.