Question in the photo

Rama09 [41]

There are a variety of waves from light waves to mechanical waves. Waves can exhibit different effects like the Doppler Effect.

All light waves behave in a similar manner. They either get transmitted, reflected, absorbed, refracted, polarized, diffracted, or scattered based off of the composition of the object and the wavelength of the light.

According to Wikipedia, “One important property of mechanical waves is that their amplitudes are measured in an unusual way, displacement divided by (reduced) wavelength. When this gets comparable to unity, significant nonlinear effects such as harmonic generation may occur, and, if large enough, may result in chaotic effects.” Mechanical waves are chaotic and its “amplitudes” are measured unusually.

Diffraction is when light bends around objects and spread after passing out through small openings. “Diffraction occurs with all waves, including sound waves, water waves, and electromagnetic waves such as light that the eye can see.”-Wikipedia. Here is the formula to Diffraction: d sin θ = nλ

Doppler effect can occur for any type of wave like sound or water waves. An example of this is when we hear a police car with its sirens on, coming towards us. The closer you are to the police car, the higher the wavelength, but the farther away you are, the lower the wavelength.

5 0

3 years ago

Electronic flash units for camera contain a capacitor for storing the energy used to produce the flash. In one such unit, the fl

nadezda [96]

Answer:

420J

Explanation:

Power is the time rate of change in energy. Power is the ratio of energy to time. The S.I unit of power is in watts.

Given that the flash lasts for 1/675 s, power output is 2.7 * 10⁵ W. Hence:

Power = Energy / time

Substituting:

2.7 * 10⁵ W = Energy / (1/675)

Energy = 2.7 * 10⁵ W * 1/675 = 400J

Therefore the energy emitted as light is 400J.

Since the conversion of electric energy to light is 95% efficient, hence the energy stored as electrical energy is:

Energy(capacitor) = 5% of 400J + 400J = 0.05*400 + 400

Energy(capacitor) = 420J

8 0

2 years ago

THIS MARCIN

nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0

3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

Classify the following elements as alkine, alkine-earth, or transition metals based on their position on the periodic table

azamat

Explanation:

Iron transition metal

Potassium Alkaline metal

Strontium Alkaline earth metal

Platinum transition metal

The periodic table arranges elements based on their atomic numbers into periods and groups. Atomic number is the number of protons an atom contains.

On the periodic table:

Group IA is called Alkaline metal

Group IIA is the Alkaline earth metals

Group IIIA Boron family elements

Group VIIA Halogens

Group O Noble gases or inert gases.

Group IIIB to IIB Transition elements.

Running a check of the given elements on the periodic table will confirm answer.

Learn more:

Periodic table brainly.com/question/8543126

#learnwithBrainly

7 0

3 years ago