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3 years ago

A myopic (nearsighted) lawyer has a far point of 60.0 cm. What is the diopter value of suitable corrective contacts?

Physics

1 answer:

Vlada [557]3 years ago

6 0

Answer:

P = -1.67 dP

Explanation:

given,

myopic lawyer has a far point of = 60.0 cm = 0.6 m

If a person is suffering from myopia then he cannot see the farthest object clearly.

Image of far object does not form on the retina so, the image appear to be blur.

using lens formula

$\dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{f}$

$-\dfrac{1}{0.6} = \dfrac{1}{f}$

$\dfrac{1}{f} =P$

$P = -\dfrac{1}{0.6}$

P = -1.67 dP

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A cell phone is released from the top with the speed of 10ms what is the speed 3s after?

sergeinik [125]

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

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3 years ago

A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start

Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4 m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction = PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno.

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3 years ago

Find a first-degree polynomial function p1 whose value and slope agree with the value and slope of f at x = c. f(x) = cot(x), c

Neporo4naja [7]

The correct answer is y=-2x+(1/2)

y = f'(x)· x + c

Y = -2x + C

1 = -2x π/4 + C

=) C = I + π/2

y=-2x+(1/2) is the first-degree polynomial.

First-degree polynomials are the simplest polynomials. Here, we'll talk about a few qualities and connect the terms polynomial, function, and equation. Write a polynomial equation in standard form before attempting to solve it. Factor it, then set each variable factor to zero after it has reached zero. The original equations' answers are the solutions to the derived equations. Factoring cannot always be used to solve polynomial equations. For instance, the polynomial 2x+5 has an exponent of 1. The most typical kinds of polynomials used in algebra and precalculus are zero polynomial functions.

Learn more about polynomial functions here :-

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2 years ago

Angela has been feeling fatigued. A test to check the basal metabolic rate revealed that she has a low metabolic rate. What is A

polet [3.4K]

Here is the answer to the given question above. If Angela has been feeling fatigued and a test is used to check the basal metabolic rate and revealed that she has a low metabolic rate, therefore, the possible diagnosis for Angela would be HYPOTHYROIDISM. <span>The BMR test works by precisely measuring the amount of oxygen that you consume when your body is basal, or completely at rest. Hope this answers your question.</span>

8 0

4 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

4 years ago