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zloy xaker [14]

2 years ago

12

The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f

rom the eye. Does the older person’s eye have a larger or smaller refractive power than a normal eye, when focused on an object at the near point? Explain.

1 answer:

tensa zangetsu [6.8K]2 years ago

4 0

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

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A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed

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Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

5 0

1 year ago

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Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa

Katarina [22]

Answer:

A 93%

Explanation:

$P_1=P_2$ = Pressure will be equal at inlet and outlet

$\rho$ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

$v_1$ = Velocity at inlet = 1.2 m/s

$v_2$ = Velocity at outlet

$r_1$ = Radius of inlet = $\dfrac{1.5}{2}=0.75\ cm$

$r_2$ = Radius of outlet

From Bernoulli's relation

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s$

From continuity equation

$A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m$

The fraction would be

$\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%$

The fraction is 93.0304%

8 0

2 years ago

A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.

Volgvan

Answer:

A=0.199

Explanation:

We are given that

Mass of spring=m=450 g= $=\frac{450}{1000}=0.45 kg$

Where 1 kg=1000 g

Frequency of oscillation= $\nu=1.2Hz$

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator= $E=\frac{1}{2}m\omega^2A^2$

Where $\omega=2\pi\nu$ =Angular frequency

A=Amplitude

$\pi=\frac{22}{7}$

Using the formula

$0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2$

$A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398$

$A=\sqrt{0.0398}=0.199$

Hence, the amplitude of oscillation=A=0.199

4 0

2 years ago

A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a

Korvikt [17]

Answer:

$v^2=v_o^2-2\times a\times (v_o.t)$

Explanation:

Given:

Initial velocity of the vehicle, $v_o$

distance between the car and the tree, $x_f$

time taken to respond to the situation, $t$

acceleration of the car after braking, $a$

Using equation of motion:

$v^2=u^2+2a.s$ ..............(1)

where:

$v=$ final velocity of the car when it hits the tree

$u=$ initial velocity of the car when the tree falls

$a=$ acceleration after the brakes are applied

$s=$ distance between the tree and the car after the brakes are applied.

$s=v_o\times t$

Now for this situation the eq. (1) becomes:

$v^2=v_o^2-2\times a\times (v_o.t)$ (negative sign is for the deceleration after the brake is applied to the car.)

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2 years ago

A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

2 years ago