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Nezavi [6.7K]
3 years ago
10

FILL IN THE BLANK. The __________ of a vector is represented by the length of the arrow.

Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

magnitude

Explanation:

because

You might be interested in
An elderly sailor is shipwrecked on a desert island but manages to save his eyeglasses. The lens for one eye has a power of 1.16
igomit [66]

Answer:

m = 8

Explanation:

A telescope is a device that allows us to see objects that were very far from us, it is built by the combination of two lenses, the one with the lowest focal length near the eye and that is the one or the one with the greatest focal length, the most eye-flounder . The magnification of the telescope is

            m = - f₀ / f_{e}

Where f₀ is the focal length of the lens and f_{e} is the false distance of the eyepiece.

It is this problem that gives us the diopter of each lens, these are related to the focal length in meters

           D = 1 / f

Let's find the focal length

       f₁ = 1 / D₁

       f₁ = 1 / 1.16

       f₁ = 0.862 m

     

        f₂ = 1 / 9.37

        f₂ = 0.1067 m

Therefore, the lens with f₂ is the eyepiece and the slow one with the  

distance focal  f₁ is the objective.

Let's calculate

       m = - f₂ / f₁

      m = - 0.862 / 0.1067

      m = 8

8 0
3 years ago
Suppose you dissolve 12.8 g of one substance in 11 g of another. Is this reasonable? Explain
RUDIKE [14]

Answer:

No, its not reasonable.

Explanation:

The substance that is to be dissolved is known as solute. The substance that is dissolving is known as solvent.

The amount of solvent  in the mixture should be greater than that of solute.

Suppose we are taking a solvent in a beaker and we are continuously adding solute in it. Initially the solute dissolve quickly. At some point the solute stops dissolving in the solvent. This is known as saturation point of the solvent. After saturation point if solute is added further it does not dissolve in the solvent.

So, its not possible to dissolve 12.8 g of one substance in 11 g of another.

5 0
3 years ago
Is a mirrors surface transparent translucent or opaque how do you know.
Eva8 [605]
There are actually two different kinds of mirrors, and the answer is different
for each one.

-- Plain old everyday hand mirror, vanity mirror, bathroom mirror, makeup
mirror, etc.

Opaque, reflecting silver coating is on the back of the glass. 
Light from your tongue or your teeth flows to the front surface of the glass,
through the glass, out of the back surface of the glass, bounces off of the silver
coating on the back, reverses its direction, enters the back surface of the glass,
comes back through the glass again, leaves the front of the glass, goes into your
eyes, and you can see your teeth or your tongue.

Both surfaces of the glass, as well as the glass in between the surfaces, are
transparent.  The silver coating on the back is opaque.  I know that, because
when I look at the back of a mirror, I can't see any light coming through it. 
The coating on the back is also reflective ... a big part of the reason why
a mirror works.

-- Expensive mirrors used by astronomers and eye-doctors.
Known as "first surface" mirrors.

Opaque, reflecting silver coating is on the <em>front</em> of the glass. 
Light from your tongue or your teeth flows toward the front surface of the glass,
but never actually gets there.  It bounces off of the silver coating on the front of
the glass, reverses its direction, goes into your eyes, and you can see your teeth
or your tongue.

The glass is transparent, but that doesn't matter, because the light never reaches
the glass. It only goes as far as the opaque silver coating on the front, and is
reflected from there.

4 0
3 years ago
in order to generate electricity, nuclear powerplants take advantage of this part of the electromagnetic spectrum
antiseptic1488 [7]
Bit of an odd question. Power Plants are known to use water-powered turbines to generate electricity, but can also make use of nuclear fission.
8 0
3 years ago
A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general
valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength \lambda_i

• The wave does not satisfy the boundary conditions y_i(0;t)=0&#10;

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions,y(0,1)=0&#10;

and y(L,t)=0

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength \lambda_i  can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, y_i(0;t)=0 and y_i(L;t)=0, that are satisfied only for particular value of \lambda_i  .

Given below are the incorrect options about the wave in the string.

•  A_i must be chosen so that the wave fits exactly o the string.

• Any one of  A_i or \lambda_i  or f_i  can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies f_i and the wavelength \lambda_i  must be such that y_i(0;t) = y_i(L;t)=0&#10;

(C)

Expression for the wavelength of the various normal modes for a string is,

\lambda_n=\frac{2L}{n} (1)

When n=1 , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

\lambda_n=\frac{2L}{1}\\\\2L

When n=2 , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

\lambda_n=\frac{2L}{2}\\\\L

When n=3, this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

\lambda_n=\frac{2L}{3}

Therefore, the three longest wavelengths are 2L,L and \frac{2L}{3}.

(D)

Expression for the frequency of the various normal modes for a string is,

f_n=\frac{v}{\lambda_n}

For the case of frequency of the i^{th} normal mode the above equation becomes.

f_i=\frac{v}{\lambda_i}

Here, f_i is the frequency of the i^{th} normal mode, v is wave speed, and \lambda_i is the wavelength of i^{th} normal mode.

Therefore, the frequency of i^{th} normal mode is  f_i=\frac{v}{\lambda_i}

.

6 0
3 years ago
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