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maria [59]
3 years ago
10

Standing on a mountain, you look towards another mountain 6.00 km away, on which there are two persons standing 1.00 meter apart

. If the pupil of your eye has a diameter of 3.50 mm, will you be able to resolve the two persons (or will you see them as one). Take wavelength of light = 600 nm
Physics
1 answer:
allsm [11]3 years ago
5 0

Answer:

No.

Explanation:

We shall solve this problem by calculating the resolving power of eye for given wavelength

Resolving Power of eye = \frac{1.22\lambda }{D}

Where λ is wave length of light and D is diameter of eye.

λ is 600 nm and D is 3.5 mm . Put these values in the given formula

Resolving Power = \frac{1.22\times 600\times 10^{-9} }{3.5\times 10^{-3}}\\

=209.14 \times 10^{-6}radian

From the formula

Φ = \frac{L}{D}[/tex]

Where Ф is resolving power . If L be distance between two points that can be resolved at distance D. D is 6 km or 6000 m .

209.14  \times 10^{-6}=\frac{L}{6000}\\

L= 1.254 m

So minimum distance that can be resolved is 1.254 m.

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3 years ago
A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide suffi
Mademuasel [1]

The image of the water tower and the houses is in the attachment.

Answer: (a) P = 245kPa;

(b) P = 173.5 kPa

Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.

The pressure is calculated as: P = hρg

where

ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;

When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:

ρ = \frac{m}{V}

V = ρ/m

V = \frac{5.25.10^{5}}{10^{3}}

V = 525 m³

To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:

V = \frac{4}{3}.\pi.r^{3}

r = \sqrt[3]{ \frac{3}{4\pi } .V}

r = \sqrt[3]{\frac{525.3}{4\pi } }

r = 5.005 m

diameter = 2*r = 10.01m

(a) Height for House A:

h = 15 + 10.01

h = 25.01

P = hρg

P = 25.01.10³.9.8

P = 245.10³ Pa or 245kPa

(b) h = 25 - 7.3

h = 17.71

P = hρg

P = 17.71.1000.9.8

P = 173.5.10³ Pa or 173.5 kPa

4 0
3 years ago
A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

y(t)=2.80t+0.61t^3

The velocity of the crate vs time is given by the derivative of the position, so it is:

v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

F(t)=ma(t)

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

a(4.10)=3.66\cdot 4.10 =15.0 m/s^2

And therefore, the force on the crate is:

F=ma=(5.00)(15.0)=75 N

7 0
3 years ago
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