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aleksandrvk [35]

4 years ago

11

A woman wears bifocal glasses with the lenses 2.0 cm in front of her eyes. The upper half of each lens has power-0.500 diopter a

nd corrects her far vision so that she can focus clearly on distant when looking through that half. The lower half of each lens has power +2.00 diopters and corrects her near vision when she looks through that half What are the far point and near point of her eyes?

1 answer:

MrRissso [65]4 years ago

4 0

Answer:

q = -2 m and q = -0.5 m

Explanation:

For this exercise we must use the equation of the optical constructor

1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and the image, respectively

Let's start with the far vision point, in this case the power of the lens is

P = -0.5D

power is defined as the inverse of the focal length in meter

f = 1 / D

f = -1 / 0.5

f = - 2m

the object for the far vision point is at infinity p = infinity

1 / f = 1 / p + i / q

1 / q = 1 / f - 1 / p

1 / q = -1/2 - 1 / ∞

q = -2 m

The sign indicates that the image is on the same side as the object

Now let's lock the near view point

D = +2.00 D

f = 1 / D

f = 0.5m

the near mink point is p = 25 cm = 0.25 m

1 / f = 1 / p + 1 / q

1 / q = 1 / f - 1 / p

1 / q = 1 / 0.5 - 1 / 0.25

1 / q = -2

q = -0.5 m

the sign indicates that the image is on the same side as the object in front of the lens

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A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.

dedylja [7]

Answer:

F = 768 N

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, $\dfrac{dm}{dt}=240\ kg/min = 4\ kg/s$

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

$F=\dfrac{dp}{dt}$

$F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$

Since, the speed is constant,

$F=m\dfrac{dv}{dt}$

$F=v\dfrac{dm}{dt}$

$F=3.2\times 240$

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0

4 years ago

Use Newton’s Universal Law of Gravitation to calculate the magnitude of the gravitational force between a 200 kg refrigerator an

expeople1 [14]

Answer:

3.735×10⁻⁶ N

Explanation:

From newton' s law of universal gravitation,

F = Gmm'/r² .............................. Equation 1

Where F = Gravitational force between the person and the refrigerator, m = mass of the person, m' = mass of the refrigerator, r = distance between the person and the refrigerator. G = gravitational universal constant.

Given: m = 70 kg, m' = 200 kg, r = 0.5 m

Constant: G = 6.67×10⁻¹¹ Nm²/kg².

F = (6.67×10⁻¹¹×70×200)/0.5²

F = 93380×10⁻¹¹/0.25

F = 373520×10⁻¹¹

F = 3.735×10⁻⁶ N

Hence the force between the person and the refrigerator = 3.735×10⁻⁶ N

6 0

4 years ago

A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill

LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

$\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s$

(ii) The linear speed of the small metal object is calculated as;

$v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s$

6 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

A 50 N object is on Earth. What is its mass?

faust18 [17]

Answer: 5.10 KG

Explanation: W = 50 N

BY FORMULA

W = MG

M = W/G

M = 50 /9.8

M= 5.10 KG

7 0

2 years ago