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Nat2105 [25]
3 years ago
12

f a wave has a wavelength of 9 meters and a period of 0.006, what is the velocity of the wave? A. 1,200 m/s B. 1,500 m/s C. 1,80

0 m/s D. 1,285 m/s
Physics
2 answers:
SVETLANKA909090 [29]3 years ago
7 0

Answer:

B

Explanation:

GalinKa [24]3 years ago
6 0

Answer: 1500 m/sec. ( b )

Solution:

V= f (λ)

Speed = frequency. Wavelength  

And frequency = 1/time period

First we will find the frequency :

Frequency = 1/time period

1/0.006 = 166.67 Hz

Using this in the equation:

Speed = frequency. Wavelength  

V= f (λ)

9 × 166.67  

= 1500.03 ≈ 1500

V (Speed) = 1500 m/s.

Note: Do not forget to mention the units or otherwise they may cost you marks!


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How fast does a 2 MeV fission neutron travel through a reactor core?
Artemon [7]

Answer:

The answer is 1.956 \times 10^7\ m/s

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy E = mc^2, so the definition of energy in this case is

E = \frac{1}{2}m v^2.

From the last equation,

v= \sqrt{2E/m}

where

E = 2 MeV = 3.204 \times 10^{-13} J

and the mass of the neutron is

m = 1.675\times 10^{-27}\ Kg.

Then

v = 1.956 \times 10^7\ m/s

the equivalent of 0.065 the speed of light.

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How many stars are in a galaxy? hundreds thousands millions billions.
faust18 [17]

Answer:

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7 0
2 years ago
Which equation can be used to model simple harmonic motion
mixer [17]

Answer:x(t)= Acos(wt)

Explanation:

According to Newton's 2nd law,a particle of mass m acted on by a force is given by:Fs=-kx

Where x is displacement from equilibrium

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Therefore X(t) = Acos(2pit/T)

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8 0
3 years ago
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
Can you explain why number 3 is correct?
lukranit [14]
No waves because Q19 waves would going at the surface at regions
7 0
2 years ago
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