1) 0.1 kWh
Explanation:
The power of the lightbulb is:
The energy used by the lightbulb is equal to the product between the power (P) and the time (t):
Since the bulb has been left on for t = 1 h, the total electrical energy used is
2) $ 0.018
Explanation:
The company charges $0.18 per each kilowatt-hour. So, we can find the total cost charged for using 0.1 kWh of electricity by setting the following proportion:
$0.18 : 1 kWh = x : 0.1 kWh
And solving for x, we find
This question is incomplete, the complete question is;
People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.
a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.) __ mm
Answer:
1) the power in D of the eyes of a woman is 61.7647 D
2) the size in mm of an image of a 8.00 mm object is -1.882 mm
Explanation:
Given the data in the question;
a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,
so
u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m
v = 2.00 cm = ( 2 / 100 )m = 0.02 m
Now, we know that power of lens p = 1 / u + 1 / v
so we substitute
p = ( 1 / 0.085 ) + ( 1 / 0.02 )
p = 11.7647 + 50
p = 61.7647 D
Therefore, the power in D of the eyes of a woman is 61.7647 D
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)
we know that;
m = -v / u
we substitute
m = -0.02 / 0.085
m = -0.2353
since H₀ = 8.0 mm
H = m × H₀
H = -0.2353 × 8.0
H = -1.882 mm
the size in mm of an image of a 8.00 mm object is -1.882 mm