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USPshnik [31]
4 years ago
13

Which wave has the greatest amplitude?

Physics
2 answers:
pychu [463]4 years ago
4 0

It is the red line the purple line carries the most the green line is the largest and the smallest is sky blue

Artyom0805 [142]4 years ago
4 0

Explanation:

The maximum displacement of the particles of wave is called its amplitude. The amplitude of purple wave is greatest i.e. wave A

The energy of wave is directly proportional to the square of amplitude. So, the purple wave carries the most energy i.e. wave A.

The number of oscillations or vibrations per unit time is called frequency of wave. The frequency of blue wave is largest i.e. wave C

The distance between two consecutive crests or two troughs is called the wavelength of wave. Wavelength and frequency have inverse relationship. So, the wavelength of blue wave is smallest i.e wave C

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You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
4 years ago
How many volts would it take to push 1 amp through a resistance of 1 ohm?
ELEN [110]
V=I x R so V= 1 x 1 =1V
5 0
3 years ago
Read 2 more answers
In order for a color to be seen by a human, the light must be
ololo11 [35]
In the part of the spectrum our eyes can detect (a spectrum is an arry of entities, as light waves or particles, ordered in accordance with the magnitudes of a common physical property, as wavelength or mass) Hope this helps you :D
8 0
3 years ago
A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp
alex41 [277]

Answer:

5 m/s2, left

Explanation:

We can solve the problem by applying Newton's second law of motion, which  states that:

\sum F=ma

where:

\sum F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have:

\sum F=30 N - 20 N = 10 N (to the left) is the net force on the object

m = 2.0 kg is the mass

So, the acceleration is:

a=\frac{\sum F}{m}=\frac{10}{2.0}=5.0 m/s^2

in the same direction as the force (left).

5 0
3 years ago
In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
4 years ago
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