What is N{2}+H{2}=NH{3}

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

2 years ago

Which one of the following statements does not accurately describe vibrations?

Ivanshal [37]

This is an example of resonance - when one object vibrating at the same natural frequency of a second object forces that second object into vibrational motion. The result of resonance is always a large vibration.

Answer D. Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body Because this statement contridicts the above statement, it is not accurate

6 0

3 years ago

Read 2 more answers

A certain lightning bolt moves 10.0 C of charge. How many charges have been moved if the fundamental unit of charge is 1.6 x 10-

Andrej [43]

To solve this problem, use the ratio given by the total number of electrons or protons that exist as a function of the total charge, and inversely proportional to the value of the fundamental charge. The number of fundamental unit $|q_e|$ that constitutes a charge of 40.0C can be calculated as

$N = \frac{|Q|}{|q_e|}$

Here,

$q_e$ = Value of charge and it is the fundamental charge

Q = Total Charge

N = Total number of electron or protons

The number of fundamental units is calculated as follows

$N = \frac{10.0C}{1.6*10^{-19}C}$

$N = 6.25*10^{19}$

Therefore the number of fundamental charge units moved by lightning bolt is $6.25*10^{19}$

8 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

2 years ago

A technician is diagnosing a car that had both front brake hoses replaced 5,000 miles ago. An inspection reveals that the right

Mamont248 [21]

Answer:

Option B

Technician B only is correct

Explanation:

Technician a is wrong because the rupture was not caused by switching DOT (Department of Transportation) 3 and DOT 4 Brake fluids. As a matter of fact both brake fluids are compatible with most vehicular systems, and to a certain extent they can be used interchangeably without any adverse effect.

Technician B on the other hand, gives a more accurate reason, since a twisted brake hose will definitely fail and rupture once enough force is applied to it, which is most likely the case of what happened considering the mileage it ran before rupturing. Within this mileage, it is very possible for the pressure on the already twisted brake hose to have been damaging it gradually before finally making it to rupture.

4 0

3 years ago