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3 years ago

11

a vehicle travels at a constant speed of 65 mph for 4 hours how far has this vehicle travelled in this time

1 answer:

Nostrana [21]3 years ago

3 0

260 miles.................

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Which sentence best describes a role of gravity in the formation of the

enyata [817]

Answer:

I think it's option D

Explanation:

I think it's option D but not so sure

8 0

2 years ago

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

If a 75 W lightbulb is 15% efficient, how many joules of light energy does the bulb produce every minute?

stiks02 [169]

Answer:

1 W = 1 J / sec Definition of watt is 1 joule / sec

So if a bulb uses 75 J / sec it must use

75 J/s * 60 sec / min = 4500 J/min energy used by bulb

If bulb is 15% efficient then the light delivered is

P = 4500 J / min * .15 = 675 J / min

4 0

2 years ago

18. If the distance from a converging lens to the object is less than the focal length of the lens, the image will be _______ th

skad [1K]

Here's a quick way to find out. Pick up your glasses, bifocals work best, and find the focal length with a flashlight against a book. If I remember right, the object should be magnified and upside down. So, A.

5 0

3 years ago

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

5 0

2 years ago