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Aloiza [94]
3 years ago
15

A woman can see clearly with her right eye only when objects are between 45 cm and 155 cm away. Prescription bifocals should hav

e what powers so that she can see distant objects clearly (upper part) and be able to read a book 25 cm away (lower part) with her right eye? Assume that the glasses will be 2.0 cm from the eye.
Physics
1 answer:
Yuri [45]3 years ago
8 0

Answer:

Explanation:

given,

A woman can see object between 45 cm and 155 cm

glasses is 2 cm from the eye

to able to read a book distance = 25 cm

For the  object distant apart

\dfrac{1}{f}= \dfrac{1}{p} +\dfrac{1}{q}

\dfrac{1}{f}= \dfrac{1}{\infty}+\dfrac{1}{-153}

\dfrac{1}{f} = - 6.53 \times 10^{-3}

hence, D = - 0.653  m

For the object at the near point

\dfrac{1}{f}= \dfrac{1}{p} +\dfrac{1}{q}

\dfrac{1}{f}= \dfrac{1}{23} +\dfrac{1}{-43}

\dfrac{1}{f}= 0.0202 cm

hence, D = 2.02 m

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Two piles of heavy boxes must be lifted onto a shelf. Each pile has 20 boxes and each box is the same size and weight. Tom and c
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A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored, how high (in m) would
motikmotik

Then the maximum height of the catapult will be 69.78 meters.

<h3>What is kinematics?</h3>

The study of motion without considering the mass and the cause of the motion. The equation of motion is given below.

v = u + at

s = ut + (1/2)at²

v² = u² + 2as

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored.

a = - 9.81 m/s²

u = 37 m/s

v = 0

s = h

Then the maximum height of the catapult will be given by the third equation.

v² = u² + 2as

0² = 37² - 2 × 9.81 × h

h = 37² / (2 × 9.81)

h = 69.78 meters

Then the maximum height of the catapult will be 69.78 meters.

More about the kinematics link is given below.

brainly.com/question/7590442

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4 0
1 year ago
What was the hikers average velocity during part c of the hike?
Brrunno [24]

Answer:

6 km/h west

Explanation:

During part c of the hike, the hiker moved 6 km west, and the time was 9.45 am to 10.45 am.

So, we have:

- displacement: d = 6 km west

- time taken: from 9.45 am to 10.45 am = 1 hour

Therefore, the average velocity is given by the ratio between displacement and time taken:

v=\frac{d}{t}=\frac{6 km}{1 h}=6 km/h

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4 years ago
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The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling frictio
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Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

F_n = \rho A (v - u)^2 sin \theta

v is the velocity of jet

Q = A V

v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}

v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0

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F_x = F_n sin 60^0

F_x = 2295 \times sin 60^0

F_x = 1987.52\ N

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

3 0
3 years ago
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