Answer:
4.875 V
Explanation:
N = 1300
diameter = 2.10 cm
radius = half of diameter = 1.05 cm
B1 = 0.130 T
B2 = 0 T
t = 12 ms
According to the law of electromagnetic induction,
![e = - N\frac{d\phi }{dt}](https://tex.z-dn.net/?f=e%20%3D%20-%20N%5Cfrac%7Bd%5Cphi%20%7D%7Bdt%7D)
Where, Ф be the magnetic flux linked with the coil
![e = - NA \frac{dB }{dt}](https://tex.z-dn.net/?f=e%20%3D%20-%20NA%20%5Cfrac%7BdB%20%7D%7Bdt%7D)
![e = -1300\times3.14\times{1.05\times 1.05\times 10^{-4}\times\frac{0-0.130}{12\times10^{-3}}=](https://tex.z-dn.net/?f=e%20%3D%20-1300%5Ctimes3.14%5Ctimes%7B1.05%5Ctimes%201.05%5Ctimes%2010%5E%7B-4%7D%5Ctimes%5Cfrac%7B0-0.130%7D%7B12%5Ctimes10%5E%7B-3%7D%7D%3D)
e = 4.875 V
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
Scopes, who has substituted for the regular biology teacher was charged on May the 5th, 1925 with teaching evolution from a chapter in George William Hunters textbook. Civic Biology: Presented in problems which described the theory of evolution... Hope this helps!
Answer:
x₁ = 0.62 m
Explanation:
In this exercise the force is electric, given by Coulomb's law
F =
This force is a vector, since the three charges are in a line we can reduce the vector sum to a scalar sum.
For the sense of force let us use that charges of the same sign repel and charges of the opposite sign attract.
∑ F = F₁₂ - F₂₃
They ask us to find the point where the summaries of the force is zero.
F₁₂ - F₂₃ = 0
F₁₂ = F₂₃
let's fix a reference system located in the first charge (more to the left), the distance between the two charges is d = 1.5 m and x is the distance to the location of the second sphere
k q₁q₂ / x² = k q₂q₃ / (d-x) ²
q₁ (d-x) ² = q₃ x²
let's solve
d² - 2 x d + x² =
x²
x² (1 -
) - 2x d + d² = 0
we substitute the values
x² (1- 4/2) - 2 1.5 x + 1.5² = 0
x² (-1) - 3.0 x + 2.25 = 0
x² + 3 x - 2.25 = 0
let's solve the quadratic equation
x = [-3 ±
] / 2
x = [-3 ± 4.24] / 2
x₁ = 0.62 m
x₂ = 3.62 m
since it indicates that the charge q₂ e places between the spheres, the correct solution is
x₁ = 0.62 m
Answer:
The smallest possibility is 0.01E-22kgm/s
Explanation:
Using
Momentum= h/4πx
= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)
= 0.01*10^-22kgm/s