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Nat2105 [25]
3 years ago
14

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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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
Olegator [25]

Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

S_0 = LT^{-4 --- dimension

S_0 \to m/s^4 --- unit

Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

4 0
3 years ago
Being overweight or obese increases the risk of:
Svetradugi [14.3K]

theanswer is b.heart disease

3 0
2 years ago
Read 2 more answers
A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
3241004551 [841]

Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

7 0
3 years ago
How is a scientific theory different from a scientific law?
Anuta_ua [19.1K]
Answer:

Scientific theory is a well-supported explanation of an occurrence of a natural phenomenon. Scientific law is a description of a natural phenomenon that is always observed to occur under specific circumstances.

Explanation:

5 0
3 years ago
A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t
Elis [28]

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

3 0
3 years ago
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