How are metal bridges built to cope with changes of temperature?
1 answer:
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Answer:
electricfield at (0,0,0) is Et = 2 k q / a²
Explanation:
For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity
For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation
E = k q / r²
Point (0,0,0)
We calculate for the charge -q which is at a distance R = a
E1 = k (-q) / a²
E1 = - kq / a²
As the test charge is positive in the field it goes to the left, attractive force
We calculate for the charge that is also at R = a
E2 = k q / a²
This field goes to the left, repulsive force
We find the total electric field
Et = E1 + E2
Et = kq / a² + kq / a²
Et = 2 k q / a²
Point (0,0, R)
We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a
E1 = k q / (R + a)²
E2 = kq / (R-a)²
Et = kq [1 / (R + a)² + 1 / (R-a)²]
Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}
Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}
If we use the condition that R> a we can despise in the patents "a"
(R² + a²) = R² (1+ a² / R²) ≈ R²
(R + a)² = R² (1 + a / R)² ≈ R²
(R- a)² = R² (1-a / R)² ≈ R²
Substituting in the total electric field
Et = kq {2 R²) / [R²R²]}
Et =kq 2 / R²
Answer:
v₁₀ = 1.90 m / s
Explanation:
In this exercise we are given the maximum height data, with energy we can know how fast the body came out
Final mechanical energy, maximum height
= U = m g h
Initial mechanical energy, in the lower part of the track
Em₀ = K = ½ m v²
Em=
½ m v² = m g h
v = √ 2gh
Now we can use the moment to find the speed with which objects collide
The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v
Initial before the crash
p₀ = M v₁₀ + 0
Final after the crash
= M v1f + m v
p₀ =
M v₁₀ = M + m v
As the shock is elastic the kinetic energy is conserved
K₀ =
½ M v₁₀² = ½ M ² + ½ m v²
Let's write the system of equations
M v₁₀ = M + m v
M v1₁₀² = M ² + m v²
We cleared v1f in the first we replaced in the second
= (M v₁₀ - mv) / M
M v₁₀² = M (M v₁₀ - mv)² / M² + m v²
M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²
v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0
2 m v₁₀ - v (m + 1) m/ M = 0
v₁₀ = v (m +1) / (2M)
Let's substitute the value of v
v1₁₀= √ (2gh) (m +1) / (2M)
Let's calculate
v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)
V₁₀ = 7.668 (2.68) / 10.82
v₁₀ = 1.90 m / s