The work done is 315 J
Explanation:
The graph is missing: find it in attachment.
The work done by a variable force is given by
![W=\int F(x) dx](https://tex.z-dn.net/?f=W%3D%5Cint%20F%28x%29%20dx)
where
F(x) is the magnitude of the force
x is the position
On a force vs position graph, the work done is equivalent to the area under the graph. Therefore, in order to find the work done as the block moves from x = 0 to x = 8.0 m, we have to calculate the area under the graph between these two points.
Each square corresponds (vertically) to 45 N, so the area of the first trapezium between x = 0 and x = 4.0 m is
![A_1 = \frac{(4+2)\cdot 2\cdot 45)}{2}=270](https://tex.z-dn.net/?f=A_1%20%3D%20%5Cfrac%7B%284%2B2%29%5Ccdot%202%5Ccdot%2045%29%7D%7B2%7D%3D270)
Then we have to find the area of the triangle between x = 6.0 m and x = 8.0 m:
![A_2 = \frac{1}{2}(2)(45)=45](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%282%29%2845%29%3D45)
Therefore, the total area (and the work done) is
![W=A_1+A_2=270+45=315 J](https://tex.z-dn.net/?f=W%3DA_1%2BA_2%3D270%2B45%3D315%20J)
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
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Answer:
The wavelength of radio waves is 2.85 m.
(B) is correct option.
Explanation:
Given that,
Frequency = 105 MHz
We need to calculate the wavelength of radio waves
Using formula of wave velocity
![v = f\times \lambda](https://tex.z-dn.net/?f=v%20%3D%20f%5Ctimes%20%5Clambda)
![\lambda=\dfrac{v}{f}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bv%7D%7Bf%7D)
Where, v = wave speed
f = frequency
= wavelength
Put the value into the formula
![\lambda=\dfrac{3\times10^{8}}{105\times10^{6}}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7B3%5Ctimes10%5E%7B8%7D%7D%7B105%5Ctimes10%5E%7B6%7D%7D)
![\lambda=2.85\ m](https://tex.z-dn.net/?f=%5Clambda%3D2.85%5C%20m)
Hence, The wavelength of radio waves is 2.85 m.
False.
Reasoning: actually the outer planets are made of gas and more
The relationship between frequency and wavelength for an electromagnetic wave is
![c=f \lambda](https://tex.z-dn.net/?f=c%3Df%20%5Clambda)
where
f is the frequency
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
is the wavelength
![c=3 \cdot 10^8 m/s](https://tex.z-dn.net/?f=c%3D3%20%5Ccdot%2010%5E8%20m%2Fs)
is the speed of light.
For the light in our problem, the frequency is
![f=1.20 \cdot 10^{13} s^{-1}](https://tex.z-dn.net/?f=f%3D1.20%20%5Ccdot%2010%5E%7B13%7D%20s%5E%7B-1%7D)
, so its wavelength is (re-arranging the previous formula)