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4 years ago

5

An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aper

ture of diameter D . Which aperture diameter gives the best resolution? D=(1/2)????r D=????r D=2????r

1 answer:

Neko [114]4 years ago

4 0

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

$\theta=\frac{1.22 \lambda}{D}$

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

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A 300 g bird is flying along at 6.0 m/s and sees a 10 g insect heading straight towards it with a speed of 30 m/s. The bird open

Gelneren [198K]

Answer:

(a): The bird speed after swallowing the insect is V= 4.83 m/s

(b): The impulse on the bird is I= 0.3 kg m/s

(c): The force between the bird and the insect is F= 20 N

Explanation:

ma= 0.3 kg

va= 6 m/s

mb= 0.01kg

vb= 30 m/s

(ma*va - mb*vb) / (ma+mb) = V

V= 4.83 m/s (a)

I= mb * vb

I= 0.3 kg m/s (b)

F*t= I

F= I/t

F= 20 N (c)

5 0

3 years ago

A solenoid of radius r 5 1.25 cm and length , 5 30.0 cm has 300 turns and carries 12.0 A. (a) Calculate the flux through the sur

Elina [12.6K]

Answer:

$\phi = 7.4 \times 10^{-6} Wb$

Explanation:

Magnetic field due to a long solenoid at the center is given by

$B = \frac{\mu_o N i}{L}$

here we know that

N = 300

L = 30 cm

i = 12 A

now magnetic field is given as

$B = \frac{(4\pi \times 10^{-7})(300)(12)}{0.30}$

$B = 0.015 T$

Now magnetic flux through the disc is given as

$\phi = B.A$

$\phi = (0.015)(\pi r^2)$

$\phi = (0.015)(\pi)(0.0125)^2$

$\phi = 7.4 \times 10^{-6} Wb$

8 0

3 years ago

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down,

Agata [3.3K]

Answer:

Explanation:

for vertical movement , time to reach the top = time to reach the hand = 2.5 s

v = u - gt

At the top , v = 0 , time t = 2.5 s

0 = u - g x 2.5

u = 2.5 x 9.8 = 24.5 m /s

velocity of throw = 24.5 m /s

So , when it passes the hand on its way down , it will have velocity equal to 24.5 m /s and it will accelerate downwards . Let its velocity down by 22 m be v

v² = u² + 2 g s

= 24.5² + 2 x 9.8 x 22

= 600.25 + 431.2

= 1031.45

v = 32.11 m /s .

7 0

3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$

$\Rightarrow q_1=5\times 10^{-8}-q_2$

The equation (ii) become:

$(5\times 10^{-8}-q_2)q_2=6\times10^{-16}$

$\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0$

$\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}$

From equation (iii)

$q_1=2\times10^{-8}, 3\times10^{-8}$

So, the magnitude of initial charges on both the sphere are $3\times10^{-8}$ Coulombs $=0.03 \mu C$ and $2\times10^{-8}$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

4 0

3 years ago

A. A person does 5415J of work in 4.5 minutes, how much power, expressed in watts, rounded to the tenths place, was generated?

iris [78.8K]

Answer: I only have the answer for A, but still, the answer is 20.07.

Explanation: Because the number ends in a 15, round to the nearest tenth which is 20. So now the number is 5420.
To get the Watts (W) you have to take 5420 and divide it by the time in seconds. 4.5 minutes in seconds is 270, so 5420 divided by 270 equals 20.07.
So 20.07 is the answer.

3 0

3 years ago