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3 years ago

What is the formula to calculate wave speed?

Physics

1 answer:

Kisachek [45]3 years ago

4 0

Answer: WaveSpeed = Wavelength x Frequency

Explanation: Wave speed is the distance a wave travels in a given amount of time, such as the number of meters it travels per second. Wave speed is related to wavelength and wave frequency by the equation. This equation can be used to calculate wave speed when wavelength and frequency are known.

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"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front

cluponka [151]

If i wouldve know i would tell you sorry

4 0

3 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

3 years ago

What property of light waves does the michelson-morley interferometer directly demonstrate?

yan [13]

The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.

8 0

3 years ago

The next four questions refer to the situation below.

Anna11 [10]

Answer:

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}, t_{out} = $\frac{D}{v_s +v_r}$

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

$v_{sg 1} = v_{sr} + v_{rg}$

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

$v_{sg1}$ = D / $t_{out}$

D = v_{sg1} t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

$v_{sg 2} = v_{sr} - v_{rg}$

$v_{sg 2}$ = D / $t_{in}$

D = v_{sg 2} t_{in}

with the distance is the same we can equalize

$v_{sg1} t_{out} = v_{sg2} t_{in}$

t_{out} = t_{in}

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}

This must be the answer since the return time is known. If you want to delete this time

t_{in}= D / $v_{sg2}$

we substitute

t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

t_{out} = $\frac{D}{v_s +v_r}$

7 0

3 years ago

Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second. It hits stationary ball B and they undergo

blagie [28]

<span>If Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second, and It hits stationary ball B and they undergo elastic collision, thus the two balls have different masses, then the following statement which is true is the statement that stated that there was no y-momentum initially.</span>

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3 years ago