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salantis [7]
3 years ago
5

The frequency of a wave is tripled, while the wave speed is held constant. What happens to the wavelength?

Physics
2 answers:
Mashcka [7]3 years ago
7 0

Wherever there are waves,

(the frequency) times (the wavelength) = the wave's speed

This says that the product of multiplying the frequency by the wavelength is always the same number.  So if one of them gets multiplied by something, the other one immediately gets divided by the same thing, in order to keep their product from changing.

If the frequency is tripled, the wavelength must be <em>divided by 3</em>.

grigory [225]3 years ago
6 0
The wavelength decreases
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ANEK [815]

Answer:

Explanation:

Check attachment for solution

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5 0
2 years ago
What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me
alexgriva [62]

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image'h_{i}' to the height of the object 'h_{o}'. It is denoted by letter 'm'.

Mathematically, it can be written as

m= \frac{h_{i}}{h_{o}}   ------------(1)

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance'd_{i}' to the object distance 'd_{o}'.

Mathematically, it can be written as

m= - \frac{d_{i}}{d_{o}}   ------------(2)

Now, from equation (1) and (2) we have,

m = \frac{h_{i}}{h_{o}}   = -  \frac{d_{i}}{d_{o}}  -----------(3)

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance (d_{o}) = + 8 mm.

Image distance (d_{i}) = + 16 mm

Object's height (h_{o}) = + 4 mm

Image's height (h_{i}) =?

Now, apply equation (3)

\frac{h_{i}}{h_{o}}   = - \frac{d_{i}}{d_{o}}

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4 0
3 years ago
A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.
Kobotan [32]

Explanation:

The given data is as follows.

           F = 5.75 \times 10^{-16} N

          q = 1.6 \times 10^{-19} C

          v = 385 m/s

       sin (63.9^{o}) = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

              B = \frac{F}{qv sin (\theta)}

                  = \frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}

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Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0
3 years ago
The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t
slamgirl [31]

Answer:

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\lambda=\dfrac{Er}{2k}

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Q=\lambda\timesL

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5 0
3 years ago
Which is an example of a solution?
vladimir1956 [14]

Answer:

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Explanation:

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