Answer:
Part a)
Work done is given as
W = -121.5 J
Part b)
Potential energy of the system is given as
![U = - \frac{5x^3}{3} + \frac{7 x^2}{2}](https://tex.z-dn.net/?f=U%20%20%3D%20-%20%5Cfrac%7B5x%5E3%7D%7B3%7D%20%2B%20%5Cfrac%7B7%20x%5E2%7D%7B2%7D)
Explanation:
As we know that work done by variable force is given as
![W = \int F. dx](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%20F.%20dx)
here we know that
![F = - 5x^2 + 7 x](https://tex.z-dn.net/?f=F%20%3D%20-%205x%5E2%20%2B%207%20x)
so we have
![W = \int (-5 x^2 + 7x) dx](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%20%28-5%20x%5E2%20%2B%207x%29%20dx)
so we have
![W = - 5x^3/3 + 7 x^2/2](https://tex.z-dn.net/?f=W%20%3D%20-%205x%5E3%2F3%20%2B%207%20x%5E2%2F2)
here we displace it from x = 2 to x = 5
so we will have
![W = -\frac{5}{3}(5^3 - 2^3) + \frac{7}{2}(5^2 - 2^2)](https://tex.z-dn.net/?f=W%20%3D%20-%5Cfrac%7B5%7D%7B3%7D%285%5E3%20-%202%5E3%29%20%2B%20%5Cfrac%7B7%7D%7B2%7D%285%5E2%20-%202%5E2%29)
![W = -195 + 73.5 = -121.5 J](https://tex.z-dn.net/?f=W%20%3D%20-195%20%2B%2073.5%20%3D%20-121.5%20J)
Part b)
As we know that change in Potential energy = work done
![U_f - U_i = -121.5](https://tex.z-dn.net/?f=U_f%20-%20U_i%20%3D%20-121.5)
![U - 0 = - \frac{5x^3}{3} + \frac{7 x^2}{2}](https://tex.z-dn.net/?f=U%20-%200%20%3D%20-%20%5Cfrac%7B5x%5E3%7D%7B3%7D%20%2B%20%5Cfrac%7B7%20x%5E2%7D%7B2%7D)
![U = - \frac{5x^3}{3} + \frac{7 x^2}{2}](https://tex.z-dn.net/?f=U%20%20%3D%20-%20%5Cfrac%7B5x%5E3%7D%7B3%7D%20%2B%20%5Cfrac%7B7%20x%5E2%7D%7B2%7D)
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Answer:
Hope it helps for you :)))))
Answer:
EMF = 187.5 volts
Explanation:
As per Faraday's law of electromagnetic induction we know that rate of change in the flux will induce EMF
so we can say
![EMF = -\frac{d\phi}{dt}](https://tex.z-dn.net/?f=EMF%20%3D%20-%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D)
![EMF = \frac{\phi_i - \phi_f}{\Delta t}](https://tex.z-dn.net/?f=EMF%20%3D%20%5Cfrac%7B%5Cphi_i%20-%20%5Cphi_f%7D%7B%5CDelta%20t%7D)
now we will have
![EMF = \frac{NBA - 0}{\Delta t}](https://tex.z-dn.net/?f=EMF%20%3D%20%5Cfrac%7BNBA%20-%200%7D%7B%5CDelta%20t%7D)
![EMF = \frac{50(1.5)(0.25)}{0.10}](https://tex.z-dn.net/?f=EMF%20%3D%20%5Cfrac%7B50%281.5%29%280.25%29%7D%7B0.10%7D)
![EMF = 187.5 Volts](https://tex.z-dn.net/?f=EMF%20%3D%20187.5%20Volts)
So the induced EMF in the coil will be 187.5 Volts
Answer : <em>Her weight is lower on Venus because the acceleration due to gravity is lower.</em>
Explanation :
Venus is also called as Earth's twin. This is because both the mass and the size of Earth and Venus are almost same. The acceleration due to gravity on earth is
while on Venus is
.
So, when Shari measure her weight on Venus she found her weight is lower on Venus. This is because the acceleration due to gravity is lower on the surface of Venus as compared to the Earth.
Since, ![w=mg](https://tex.z-dn.net/?f=w%3Dmg)
i.e. weight depends on g.
<em>So, correct prediction is (b)</em>