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gizmo_the_mogwai [7]
3 years ago
5

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resol

ved if their angular separation is 3.0×10−5 radians? Express your answer in nano.
Physics
1 answer:
Alex3 years ago
6 0

This question is incomplete.Here is complete one

Consider a telescope with a small circular aperture of diameter 2.0 centimeters. If two point sources of light are being imaged by this telescope, what is the maximum wavelength (lambda) at which the two can be resolved if their angular separation is 3.0x10^-5 radians? Express your answer in nanometers

Answer:

λ =492nm

Explanation:

Given Data

diameter=2.0 cm

angular separation=3.0x10^-5 radians

λ=?

Solution

sin(theta) = 1.22 x λ /D

λ=(0.02×sin(3.0×10⁻⁵))/1.22

λ=492nm

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A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet
Ede4ka [16]

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.  

7 0
3 years ago
A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m twice every second, what is the tension i
Svetradugi [14.3K]

Answer:

T = 19.75 N

Explanation:

given,

mass of ball = 0.25 Kg

radius = 0.5 m

frequency = 2 s⁻¹

tension in the string = ?

angular velocity

ω = 2 π f

ω = 2 π x 2

ω = 12.57 rad/s

tension on the string is equal to the centripetal force

T = m ω² r

T = 0.25 x 12.57² x 0.5

T = 19.75 N

Tension in the string is equal to T = 19.75 N

3 0
3 years ago
A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc
amid [387]
A= f/m

a= 19/2

a= 9.5m/s^2
8 0
3 years ago
The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

5 0
3 years ago
The bending of waves such as sound waves, light waves, and waves in water, around obstacles or the edges of openings is called.
kolezko [41]

Answer:

Diffraction

Explanation:

Diffraction is the bending of waves around obstacles and openings. The amount of diffraction increases with increasing wavelength.

4 0
11 months ago
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