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A nearsighted person has a far point of 40cm. What power spectacle lens is needed if the lens is 2cm from the eye

Answers

Answer 1

Answer:

The value is $p = - 2.63 \ Diopters$

Explanation:

From the question we are told that

The value of the far point is $a = 40 \ cm = 0.4 \ m$

The distance of the lens to the eye is $b = 2 \ cm = 0.02$

Generally

$1 Diopter = > 1 m^(-1)$

Generally the power spectacle lens needed is mathematically represented as

$p = (1)/(d_o ) + (1)/(d_i)$

Here $d_o$ is the object distance which for a near sighted person is $d_o = \infty$

And $d_i$ is the image distance which is evaluated as

$d_i = b - a$

=> $d_i = 0.02 - 0.4$

=> $d_i = -0.38 \ m$

$p = (1)/(\infty ) + (1)/(-0.38)$

=> $p = 0 - 2.63$

=> $p = - 2.63 \ Diopters$

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A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest. The two blocks stick together and move with a speed of vo /3. In terms of m, mass M is most nearly_____.

Answers

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as

$m_1 = m$

$m_2 = M$

$v_(1i) =v_0$

$v_(2i) = 0$

Initial momentum will be

$P_i = m_iv_(1i)+m_2v_(2i)$

$P_i = mv_0$

After collision

$v_(1f) = v_(2f) = (v_0)/(3)$

Final momentum

$P_f = (m_1+m_2)((v_0)/(3))$

$P_f = (m+M)((v_0)/(3))$

From conservation of momentum

$P_f = P_i$

Replacing,

$(m+M)((v_0)/(3))=mv_0$

$(m+M)(1)/(3) = m$

$m+M=3m$

$M=3m-m$

$M=2m$

A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the projectile a freely falling body?Yes or No(b) What is its acceleration in the vertical direction? (Let up be the positive direction.)____? m/s2(c) What is its acceleration in the horizontal direction?

Answers

Answer:

A) No

B)-9,81 m/s^2

C)0 m/s^2

Explanation:

A free fallin object has only velocity on the vertical axis so any object that is moving in the Y and X axis has projetile motion not free falling, and when dealing with projectile motion the object is experiencing acceleration towards the ground of -9,81m/s^2 and in the Y axis, in the X axis there´s is only acceleration if the air is providing resistance, since it states that it isnot, then the accleration is 0.

A physics cart has a projectile launcher mounted on top. While traveling on a straight track at 0.500 m/s, a projectile is fired. It lands back in the same place on top of the launcher after the cart has moved a distance of 2.30 m. In the frame of reference of the cart, (a) at what angle was the projectile fired and (b) what was the initial velocity of the projectile? (c) What is the shape of the projectile as seen by an observer on the cart? A physics student is watching the demonstration from a classroom seat. According to the student, (d) what is the shape of the projectile’s path, and (e) what is its initial velocity?

Answers

Answer:

(a) $90^(\circ)$

(b) Initial velocity of the projectile is 22.54 m/s

(d) Parabolic

(e) The initial velocity is 23.04 m/s

Solution:

As per the question:

Velocity of the cart, v = 0.500 m/s

Distance moved by the cart, d = 2.30 m

Now,

(a) The projectile must be fired at an angle of $90^(\circ)$ so that it mounts on the top of the cart moving with constant velocity.

(b) Now, for initial velocity, u':

Time of flight is given by;

$T = (D)/(v)$ (1)

where

T = Flight time

D = Distance covered

(b) The component of velocity w.r.t an observer:

Horizontal component, $v_(x) = u'cos\theta$

Vertical component, $v_(y) = u'sin\theta - gT$

Also, the vertical component of velocity at maximum height is zero, $v_(y) = 0$

Therefore, $T = (u')/(g)$

Total flight time, $(2u')/(g)$ (2)

Now, from eqn (1) and (2):

$u' = (gD)/(2v)$

$u' = (9.8* 2.30)/(2* 0.500) = 22.54 m/s$

(d) The shape of the path of the projectile seen by the physics student outside the reference frame of the cart is parabolic

(e) The initial velocity is given by:

u = u' + v = 22.54 + 0.5 = 23.04 m/s

The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.

Answers

Answer:

use google to find answer

Explanation:

69 69 69 69 69 69

Energy can be transferred from a closed system to the surroundings by: (A) Internal chemical reactions (B) Heat (C) Shaft work (D) Change in pressure without changing volume (E) Mass transfer

Answers

Answer:

option the correct is B

Explanation:

Let's analyze the different options, for a closed system

- an internal reaction changes the system, but does not affect the surrounding environment

- Heat, is a means of transfer that occurs when two bodies are in contact, one of the body can be a closed system since the only thing that happens is thermal transfer, without movement of the system itself. This is the correct result.

- Work implies a movement whereby the system must be mobile, it is not an option

- Pressure change. change in the system, but does not affect the environment

- Mass transfer is not possible in a closed system

After analyzing each option the correct one in B

In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits? (The central bright fringe is zeroth one).

Answers

The wavelength of the light passing through the slit is 214 nm.

What is the wavelength?

The wavelength is the distance between identical points in the adjacent cycles of a waveform.

Given that the separation between two slits d is 3.00 × 10^-5 m and the distance from the slit to screen r is 2 m. The distance from the central spot to fringe s is 10.0 m and the bright bands of the spectrum m are 7 for the seventh bright fringe.

The wavelength of the light passing through the slit is calculated as given below.

$\lambda = \frac {sd}{mr}$

$\lambda = \frac {10* 10^(-2) * 3.00 * 10^(-5)}{7 * 2}$

$\lambda = 2.14 * 10^(-7)\;\rm m$

$\lambda = 214 \;\rm nm$

Hence we can conclude that the wavelength of the light passing through the slit is 214 nm.

To know more about the wavelength, follow the link given below.

brainly.com/question/7143261.

To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.

By definition in the principle of superposition, light interference is defined as

$Y = (m\lambda R)/(d)$

Where,

d = Separation of the two slits

$\lambda = Wavelength$

R = Distance from slit to screen

m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.

Y = Distance from central spot to fringe.

Re-arrange the equation to find \lambda we have that

$\lambda = (Yd)/(mR)$