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vredina [299]
2 years ago
13

If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the

person would need to be able to read a book held at 0.350 m from the person's eyes? Express your answer in meters. View Available Hint(s) f2 = nothing
Physics
1 answer:
Doss [256]2 years ago
3 0

Answer:

0.84 cm

Explanation:

u = Object distance =  0.35 cm

v = Image distance = -0.6 cm (near point is considered as image distance and negative due to sign convention)

f = Focal length

From lens equation

\frac{1}{f_2}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f_2}=\frac{1}{0.35}+\frac{1}{-0.6}\\\Rightarrow \frac{1}{f_2}=\frac{25}{21}\\\Rightarrow f_2=\frac{21}{25}=0.84\ cm

Focal length of the lens is 0.84 cm

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aksik [14]

Answer:

It is showing the wavelength.

Explanation: Hope it helps you:)))

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7 0
2 years ago
What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p
irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

  • <em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
  • <em>Distance between each floor, d = 10 ft</em>

The vertical pressure of the water is calculated as follows;

P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10  ft

P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

3 0
2 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

4 0
3 years ago
Which of these IS an example of balanced forces? A. a book on a shelf B. air rushing out of a balloon C. rolling over and fallin
babunello [35]

Forces are balanced when net force on the object is zero or the sum of all force on the object is zero.

For book kept on a shelf, the weight of the book in down direction is balanced by the normal force on the book by the shelf. hence the book kept on a shelf is an example of balanced force.

In case of air rushing out of balloon , the balloon experience a net force by the air coming out of it.

rolling over and falling off the bed , the object falls under gravity force.

a car speeding up accelerates. since it accelerates , it has net force on it.

3 0
3 years ago
At which position would a person on earth see the entire lighted half of the moon?
Gemiola [76]

The best position for the person would be outside, under a clear sky, standing up.  He should do it sometime between sunset and sunrise, from a day before until a day after the moment of Full Moon.

3 0
3 years ago
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