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After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc

lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

$P = -0.5 dioptre$

now we have

$f = \frac{1}{P}$

$f = -2 m$

$f = -200 cm$

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + 0 = \frac{1}{200}$

$d_i = 200 cm$

so his far point according to this pair of glass is 200 cm

7 0

3 years ago

N in-ground swimming pool has the dimensions shown in the drawing. It is filled with water to a uniform depth of 3.00 m. The den

Doss [256]

Answer:

The pressure is $P = 1.31*10^{5} \ Pa$

Explanation:

From the question we are told that

The depth of the swimming pool is $d = 3.00 \ m$

The density of water is $\rho = 1.00*10^{3} \ kg /m^3$

Generally the total pressure exerted on the bottom of the swimming pool is mathematically represented as

$P = P_o + \rho * g * h$

Here $P_o$ is the atmospheric pressure with value

$P_o = 101325 \ Pa$

So

$P = 101325 + [1000 * 9.8 * 3]$

=> $P = 130725 \ Pa$

=> $P = 1.31*10^{5} \ Pa$

7 0

3 years ago

Electrons are ejected from a metallic surface with speeds of up to 4.60 3 105 m/s when light with a wavelength of 625 nm is used

Darya [45]

Complete question:

Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

Answer:

Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV

Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz

Explanation:

The kinetic energy (KE) of the emitted photon:

KE = 0.5mv²

m is mass of electron = 9.1 X 10⁻³¹ kg

KE = 0.5 * 9.1 X 10⁻³¹ * (460000)² = 9.628 X 10⁻²⁰ J

in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV

The photon energy of the incoming radiation:

E = hf = hc/λ

c is speed of light (photon) = 3 x 10⁸

h is Planck's constant = 6.626 × 10⁻³⁴ J.s

E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)

E = 31.805 X 10⁻²⁰ J

in eV = 31.805 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.985 eV

Part (a) the work function of the surface

KE = hf - W

where;

W is work function

W = hf - KE

W = 31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J

in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV

Part(b) the cutoff frequency for this surface

W =hf

f = W/h

f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)

f = 3.347 x 10¹⁴ Hz

8 0

3 years ago

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball

jonny [76]

(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s

(b) The work that the racket does on the ball is -35.1 Joule

<h3>Further Explanation</h3>

<u>Given :</u>

mass of ball = m = 0.06 kg

initial velocity = v₁ = -50.4 m/s

final velocity = v₂ = 37.0 m/s

<u>Unknown :</u>

(a) Impulse = I = ?

(b) Work = W = ?

<u>Solution :</u>

<h2>Question (a) :</h2>

In this question , we could use the formula from Second Law of Newton :

$I = \Delta p$

$I = p_2 - p_1$

$I = m \times v_2 - m \times v_1$

$I = m \times (v_2 - v_1)$

$I = 0.06 \times (37.0 - (-50.4))$

$I = 0.06 \times (87.4)$

$I = 5.244~kg.m/s$

$\large { \boxed {I \approx 5.24~kg.m/s} }$

<h2>Question (b) :</h2>

$W = F \times d$

$W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)$

$W = \frac{I(v_1 + v_2)}{2}$

$W = \frac{5.244(-50.4 + 37)}{2}$

$W = \frac{5.244(-13.4)}{2}$

$W = -35.1348~Joule$

$\large { \boxed {W \approx -35.1~Joule}}$

<h3>Learn more</h3>

Newton's Law of Motion: brainly.com/question/10431582

Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Newton, Law, Impulse, Work

6 0

4 years ago

Pairs of magnets are shown in the diagram.

belka [17]

The answer is X. I hope this helped

8 0

4 years ago

Read 2 more answers

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