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vredina [299]
3 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]3 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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a rock is dropped from a height of 80 m and is in free fall what is the velocity of as it reaches the ground 4.0 seconds later
Marta_Voda [28]
Velocity = displacement (distance)/time

v=80m/4s

v=20m/s

velocity = 20 meters per second
8 0
3 years ago
The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o
LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

\bar F \cdot \Delta t = m \cdot  (v_{2}-v_{1}) (1)

Where:

\bar F - Average impact force, in newtons.

\Delta t - Duration of the impact, in seconds.

m - Mass of the sledge hammer, in kilograms.

v_{1}, v_{2} - Initial and final velocity, in meters per second.

If we know that \Delta t = 0.0020\,s, m = 4\,kg, v_{1} = -6\,\frac{m}{s} and v_{2} = 0\,\frac{m}{s}, then we estimate the average impact force is:

\bar F = \frac{m\cdot  (v_{2}-v_{1})}{\Delta t}

\bar F = 12000\,N

The average impact force is 12000 newtons.

5 0
3 years ago
Hi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.
kifflom [539]

Answer:

The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.

5 0
3 years ago
Three conducting plates, each of area A, are connected as shown.
Shkiper50 [21]
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two. 
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂) 
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) ) 
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂ 
But I suppose we ought to kick that idea around a bit. 
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D. 
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁² 
Differentiate with respect to d₁ 
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero. 
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D 
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that 
d₁ = d₂ = ½D so 
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
7 0
3 years ago
for an object that is speeding up in the positive direction, what does the displacement vs. time graph look like?
slava [35]

Answer:

The speeding up is steady, it is a parabola (a=V*t+(at^2)/2), and give it's an equation in connection to time, at that point, it is conceivable to discover the separation recipe by utilizing more substantial amount mathematics(integrals).

Explanation:

7 0
3 years ago
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