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romanna [79]
3 years ago
12

i need help with queston like right now. Each molecule of iron oxide, Fe2O3, contains ____________________ atom(s) of iron.

Physics
2 answers:
ozzi3 years ago
8 0
There are 3 iron atoms per molecule of iron oxide.
azamat3 years ago
3 0
  I need help with queston like right now. Each molecule of iron oxide, Fe2O3, contains  3 iron atoms per molecule of iron oxide.
You might be interested in
The amplitude of a simple harmonic oscillator will be doubled by:a) doubling only the initial speedb) doubling the initial displ
OLEGan [10]

Answer:

When both initial speed and initial displacement is doubled then amplitude will be doubled.

Explanation:

Given that :- Amplitude of simple harmonic Oscillator  is doubled.

So,

     Formula of Simple harmonic oscillator is  X=A\sin\ (2\pi ft +\phi)  ...........(1)

                                                            Where X = Position in (m,cm,km.....)

                                                                        A = Amplitude  in (m,cm,km.....)

                                                                        F = Frequency in (Hz)

                                                                        T = Time in (sec.)

                                                                        Ф = Phase in (rad)

  For initial displacement taking t=0 we get,

                          Initial displacement = A\sin(\phi)    .................(2)            

Now taking equation (1) and differentiating it w.r.t to (t) we get

                                \frac{dx}{dt} = 2\pi fA\cos\ (2\pi ft+\phi)

                                 V= 2\pi fA\cos\ (2\pi ft+\phi)

taking t=0 for initial speed then we get,

                                Initial speed = 2\pi fA\cos\phi    ...............(3)

observing equation (2) & (3) that the initial displacement and initial speed depends on the Amplitude of the Oscillator.

Hence,

when both initial speed and displacement is doubled then amplitude will be doubled.

4 0
3 years ago
The ______ is the component of communication through which information is transferred.
MrMuchimi
The medium is the component of communication through which information is transferred. The correct option in the given question is option "c". Medium is actually the means through which communication is done. It can be a an oral message or written message or it can be nonverbal or symbolic. Any kind of message transmission requires a medium and without an medium no message can be transmitted. Whenever people talk on the telephone or write a letter, there has to be a medium for transmitting the message to the person on the other side.
3 0
3 years ago
A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in
VikaD [51]

Answer:

B. - 0.328

Explanation

Potential Energy:<em> This is the energy of a body due to position.</em>

<em>The S.I unit of potential energy is Joules (J).</em>

<em>It can be expressed mathematically as</em>

<em>Ep = mgh........................... Equation 1</em>

<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>

<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>

Substituting these values into equation 1

Ep = 0.00274×12.2×9.8

Ep = 0.328 J.

Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J

The right option is B. - 0.328

<em />

7 0
3 years ago
2. A 55 kg woman has a momentum of 200 kg m/s. What is her velocity?
NeTakaya

Answer:

\boxed {\tt 3.63636364 \ m/s}

Explanation:

Velocity can be found using the following formula:

v=\frac{p}{m}

where p is the momentum and m is the mass.

The woman has a mass of 55 kilograms and a momentum of 200 kilogram meters per second.

p= 200 \ kgm/s\\m=55 \ kg

Substitute the values into the formula.

v=\frac{200 \ kg m/s}{55 \ kg}

Divide. Note that the kilograms, or kg, will cancel each other out.

v=\frac{200 \ m/s}{55}

v= 3.63636364 \ m/s

The woman's velocity is 3.63636364 meters per second.

6 0
3 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

h

′

=

u

v

Thus,

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

h

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
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