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guajiro [1.7K]
2 years ago
5

What real world measurable property of waves is the A value in the equation relating frequency and wavelength?

Physics
1 answer:
navik [9.2K]2 years ago
3 0

Answer:

Hello There!!

Explanation:

The frequency of a wave is the number of times per second that the wave cycles. Frequency is measured in Hertz or cycles per second.The period and frequency are closely related to each other. The period equals 1 over the frequency and the frequency is equal to one over the period.The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.

hope it helps,have a great day!!

~Pinky~

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At a certain temperature, the rate constant for this reaction is 5.64×10−4 s−1 . Calculate the half-life of cyclopropane at this
oee [108]

Answer:

t_{1/2}= 20.47\ minute

Explanation:

given,

constant rate for the reaction(k) =  5.64  ×  10⁻⁴ s⁻¹

to calculate the half life  = ?

t_{1/2} = \dfrac{0.693}{k}

t_{1/2} = \dfrac{0.693}{5.64\times 10^{-4}}

on solving the above equation

t_{1/2} = 1228.72 s

t_{1/2} = \dfrac{1228.72}{60}

t_{1/2}= 20.47\ minute

hence, the half-life of cyclopropane at this temperature =t_{1/2}= 20.47\ minute

5 0
2 years ago
A car generates 2,552 N and weighs 2,250 pounds. what is the rate of acceleration
Lilit [14]
Ans: a = 2.50 m/s^2

Explanation:
First convert the mass in its standard unit i.e. kilogram(kg):
2250 lbs = 1020.583kg

Next use Newton's Second law:
F = ma

Where F = 2552N
m = 1020.583kg

=> a = (2552/1020.583)
a = 2.50 m/s^2
6 0
2 years ago
A 1.5 kilogram car is moving at 10 meters per second east. A braking force acts on the car for 5.0 seconds, reducing its velocit
Arada [10]

Answer:

I don't know

Explanation:

8 0
3 years ago
The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/
velikii [3]

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

 To get the  true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

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ground speed=||w⃗ +P⃗ ||

 

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯,   P⃗ =AC¯¯¯¯¯¯¯¯

 

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector.  To get the coordinates of these two vectors,  use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

 

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8 0
2 years ago
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Damm [24]

Answer:

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7 0
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