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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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Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and
Sindrei [870]

Answer: a) v = ω /k, b) v = - ωAcos( kx −ωt)

Explanation:

y(x,t)=Asin(kx−ωt) defines the wave equation.

a)

We are asked to find wave speed (v)

Recall that v = fλ

From the wave equation above,

k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k

ω = 2πf where f is the frequency and ω is the angular frequency.

f = ω/ 2π.

By substituting for λ and ω into the wave speed formulae, we have that

v =( ω/ 2π) × (2π /k)

v = ω/k

b)

y(x,t)=Asin(kx−ωt)

The first derivative of y with respect to x give the velocity (vy)

By using chain rule, we have that

v = dy/dt = A cos( kx −ωt) × (−ω)

v = - ωAcos( kx −ωt)

3 0
4 years ago
under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on
d1i1m1o1n [39]
In a non accelerated frame, also called inertial frame, when no additional forces are acting upon the body.

For instance, a scale in free fall will not record the weight, but zero, like the astronauts. But also, if someone is pulling the object or pushing it, the scale will record some other value.
4 0
3 years ago
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A proton has been accelerated from rest through a potential difference of -1350 V. What is the proton's kinetic energy, in elect
Luba_88 [7]

Answer:

1 eV = 1.60 * 10^-19 J      work done in accelerating electron thru 1 V

KE (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules

1/2 m v^2 = KE = 2.16 * 10^-16 J

v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11

v = 5.09 * 10^5 m/s

8 0
3 years ago
Two different simple harmonic oscillators have the same natural frequency (f=8.80 Hz) when they are on the surface of the Earth.
bekas [8.4K]

Answer:

8.80 Hz

Explanation:

The frequency of a loaded spring is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.

Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.

On the other hand, the frequency of the simple pendulum is affected because it is given by

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by

f_2 = 8.80\sqrt{\dfrac{1.67}{9.81}}=3.63 \text{ Hz}

3 0
3 years ago
3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La
jok3333 [9.3K]

Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}

f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}    ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz

Hence, the first harmonic has a frequency of 12.90 Hz

4 0
3 years ago
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