Carol is farsighted ( presbyopia) and cannot see objects clearly that are closer to her eyes than about meter. She sees objects clearly with a relaxed eye when they are distant. What is the refractive power of reading glasses that would allow her to read a book 50 cm away with a relaxed eye

Answers

Answer 1

Answer:

1.0 dioptres

Explanation:

Farsightedness is an eye defect in which a person can see far objects clearly but not near objects. That implies that the patients' near point is farther than 25cm which is the normal least distance of distinct vision.

Farsightedness results from the eyeball being too long or the crystalline lens not being sufficiently converging.

Carol is farsighted with a near point of about a meter (100cm). We desire to make a lens to enable her near point be reduced to about 50cm. The focal length and power of this lens is calculated in the image attached.

The power of a lens is the inverse of its focal length in meters hence the 100 in the formula for power of the lens.

Answer 2

Answer:

Answer:

+1.00 diopter

Explanation:

The power of a lens can be described simply as the reciprocal of the focal length of the lens measured in meters.

But f is unknown, hence we look for the focal length with the formula

1/f = 1/u + 1/v

where u is former near point = 100cm

v is the new intended near point = 50cm

1/f = 1/50 - 1/100

1/f = 1/100

f = 100 cm

Hence we get Power (D) = 1/f

where f = focal length of the lens in meter

From the question, the focal length of the lens = 100cm = 1m

Hence D = 1/1

D = +1.00

Hence the refractive power of the reading glasses that would allow Carol to read a book 50cm away from the relaxed eye will be +1.00 diopters.

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When a box is placed on an inclined surface with no friction, it will:
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Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?
A 133 kg horizontal platform is a uniform disk of radius 1.95 m and can rotate about the vertical axis through its center. A 62.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 28.5 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

Light from a sodium vapor lamp (λ-589 nm) forms an interference pattern on a screen 0.91 m from a pair of slits in a double-slit experiment. The bright fringes near the center of the pattern are 0.19 cm apart. Determine the separation between the slits. Assume the small-angle approximation is valid here.

Answers

Answer:

separation between the slits is 0.28 mm

Explanation:

given data

wave length λ = 589 nm = 589 × $10^(-9)$ m

distance between slits and the screen D = 0.91 m

fringes weight y = 0.19 cm = 0.19 × $10^(-2)$ m

solution

we find here the spacing between the two slits i.e d

so use here formula that is

y = λD ÷ d .........................1

put here value we get

0.19 × $10^(-2)$ = $(589*10^(-9)*0.91)/(d)$

solve we get

d = 0.28 mm

An 800-kHz sinusoidal radio signal is detected at a point 6.6 km from the transmitter tower. The electric field amplitude of the signal at that point is 0.780 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the amplitude of the magnetic field of the signal at that point

Answers

Answer:

0.0000000026 T

Explanation:

$E_0$ = Maximum electric field strength = 0.78 V/m

$B_0$ = Maximum magnetic field strength

c = Speed of light = $3* 10^8\ m/s$

Relation between amplitudes of electric and magnetic fields is given by

$E_0=B_0c\n\Rightarrow B_0=(E_0)/(c)\n\Rightarrow B_0=(0.78)/(3* 10^8)\n\Rightarrow B_0=0.0000000026\ T$

The amplitude of the magnetic field is 0.0000000026 T

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.

Answers

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_(o)+a\cdot t$ ), we expand the previous expression:

$-f = \left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_(o)$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,(m)/(s^(2))$ , $v_(o) = 1.37\,(m)/(s)$ , $v = 0\,(m)/(s)$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left((52.4\,N)/(9.807\,(m)/(s^(2)) ) \right)\cdot \left((0\,(m)/(s)-1.37\,(m)/(s) )/(2.8\,s) \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

Final answer:

The magnitude of the friction force acting on the box is determined by calculating the box's acceleration, establishing its mass based on its weight information, and applying these values in Newton's second law. The calculated value is 2.62 N.

Explanation:

To determine the magnitude of the friction force, we first have to compute the acceleration of the box. Acceleration (a) can be found using the formula 'final velocity - initial velocity / time'. Since the final velocity is 0 (the box stops), and the initial velocity is 1.37 m/s, and the time is 2.8 s, we get: a = (0 - 1.37) / 2.8 = -0.49 m/s^2. The negative sign indicates deceleration.

Next, we use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration. The net force in this case is the frictional force because there is no other force acting on the box in the horizontal direction. However, we do not know the mass of the box, but we do know its weight, and weight = mass x gravitational acceleration (g). So mass = weight/g = 52.4N / 9.8m/s^2 = 5.35 kg.

Lastly, we substitute the mass and deceleration into Newton's second law to find the frictional force (f): f = mass x deceleration = 5.35kg x -0.49m/s^2 = -2.62 N. Again, the negative sign indicates that the force acts opposite to the direction of motion. Thus, the frictional force magnitude is 2.62 N.

Learn more about Friction force here:

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An iceskater is turning at a PERIOD of (1/3) second with his arms outstretched. a) What is his ANGULAR VELOCITY w? b) If he pulls his arms towards his body to reduce his MOMENT OF INTERTIA by 1/2, what is his ANGULAR VELOCITY w? c) How much does his ROTATIONAL KINETIC ENERGY change? That is, if the initial Kinetic Energy is (KE)initial, what is the final KE? d) Where did that ENERGY come from, or go to?

Answers

Answer:

Explanation:

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final = 1/2 I₂ ω₂²

= 1/2 (I₁ / 2) (2ω)²

= I₁ ω²

c )

Change in rotational kinetic energy

= I₁ ω² - 1/2 I₁ ω²

= + 1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

Tony brought 9 2/3pitchers of juice to a volleyball game, and the players drank3 7/8pitchers of it. How much juice is left?

Answers

Rewrite the amounts as improper fractions:

9 2/3 = 29/3

3 7/8 = 31/8

Rewrite both fractions with a common denominator

29/3 = 232/24

31/8 = 93/24

Now subtract: 232/24 - 93/24 = 139/24

Rewrite as a proper fraction: 5 19/24

Answer 5 19/24

Feest. Fysics and motion11
Select the correct answer
You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same point you started
from, what is your average velocity?
A
0 kilometers/hour
B.
2 kilometers/hour
4 kilometers/hour
D
8 kilometers/hour
E.
16 kilometers/hour
Rese

Answers

Velocity depends on the straight-line distance between your start-point and your end-point, regardless of what route you follow to get there.

If you stop at the same point where you started, then that distance is zero, no matter how far you drove before you returned to your start-point.

So the average velocity around any "CLOSED" path is zero. (A)

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A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)