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2 years ago

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Easy Physics question plz help ASAP

1 answer:

Aleks04 [339]2 years ago

5 0

Answer:

B___________________________________________:)

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I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

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

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

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

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

3 0

3 years ago

The law of inertia applies to objects

Anestetic [448]

Answer:

<em>at</em><em> </em><em>rest</em><em> </em><em>and</em><em> </em><em>in</em><em> </em><em>motion</em>

Explanation:

<em>The</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>inertia</em><em> </em><em>applies</em><em> </em><em>to</em><em> </em><em>objects</em><em> </em><em>at</em><em> </em><em>rest</em><em> </em><em>and</em><em> </em><em>in</em><em> </em><em>motion</em>

6 0

3 years ago

Read 2 more answers

A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is

neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed $v_{r}$ =2 m/s

velocity of motorboat relative to water is $v_{m/r}$ =4.2 m/s

so speed will be:

a) $v_{m}$ = $v_{r}$ + $v_{m/r}$

solving graphically

$v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}$

=4.7 m/s

Ф= $tan^{-1} (\frac{v_{r}}{v_{m/r}} )$

=25.5°(south of east)

b) time to cross the river: t= $\frac{w}{v_{m/r}}$ = $\frac{500}{4.2}$ =119 s

c) d= $v_{r}t$ =(2)(119)=238 m

note :

pic is attached

6 0

3 years ago

Please help me! Uniform acceleration problem sheet:

Brrunno [24]

From the calculation, the value of the acceleration is 5.8 m/s^2.

<h3>What is uniform acceleration?</h3>

The term uniform acceleration refers to a situation in which the velocity increases by equal amounts in equal time intervals.

Given the fact that the car started from rest and reached a velocity of 780.34 mph or 348.84 m/s in 1 minute of 60 seconds;

v = u + at

a = v/t

a = 348.84 m/s/ 60 seconds

a = 5.8 m/s^2

Learn more about acceleration:brainly.com/question/12550364?

#SPJ1

7 0

2 years ago

(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation

Musya8 [376]

Answer:

the answee is

2.968456 ×10^7

4 0

3 years ago