Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo
1 answer:
Answer:
Tension T1 is less than tension T2.
T1 < T2
Explanation:
According to given data,
mass of box A ( mA) is grater than mass of box B (mB)
we can write,
m(A) > m(B)
Newton's second law states that:
Tension of object is directly proportional to the mass of the system.
T ∝ m
here Boxes A and B are being pulled to the right on a frictionless surface,
so Tension T1 generates due to the mass of box A m(A)
and Tension T2 arises due to mass of the system m(A) + m(B)
Thus tension T1 will be less than tension T2
T1 < T2
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Answer:
1) E = 0 , 2) zero, 3) E = - 2,162 10⁴ N/C , 4) directed towards the center of the sphere , 5) E = 1.412 10⁴ N / C , 6)direction coming out of the sphere
Explanation:
The electric field is a vector quantity, therefore we can calculate the field due to each charge distribution and add vector
E = E₁ + E₂
To calculate each field we can use Gauss's law, which states that the flow is equal to the charge by the Gaussian surface divided by ε₀
For this case, let's take a sphere as a Gaussian surface
Ф = ∫E dA = /ε₀
The area of a sphere is
A = 4π r²
E = 1 / 4πε₀ q_{int} / r²
1) r = 4 cm
This radius is smaller than the radius of the sphere, therefore the charge inside is zero and therefore the field is zero
E = 0
2) there is no field
3) r = 8 cm
Let's calculate each field, for the inner surface
This radius is larger than the internal radius, so the field is
σ = q_{int} / A
The area of the sphere is
V = 4 π R_in²
Rho = q_{int} / 4π R_in²
q_{int} = ρ 4π R_in²
E₁ = 1 /ε₀ ρ r_in² / r²
For the outer surface
This radius is smaller so there is no load inside the Gaussian surface and therefore the field is zero
E₂ = 0
Total E
E = E₁ + 0
E = 1 /ε₀ ρ₁ R_in² / r²
Let's calculate
E = 1 /8,854 10⁻¹² 250 10⁻⁹ (7/8)²
E = - 2,162 10⁴ N / C
4) as the electric field is negative, it is directed towards the center of the sphere
5) r = 12 cm
In this case the two surfaces contribute to the electric field,
Inner surface
Q₁ = ρ₁ 4π R_in²
E₁ = 1 /ε₀ ₁rho1 R_in² / r²
Outer surface
Q₂ = ρ₂ 4π R_out²
E₂ = 1 /ε₀ ρ₂ R_out² / r²
The total field is
E = E₁ + E₂
E = 1 /ε₀ | ρ | [- R_in² + R_out²] / r²
Let's calculate
E = 1 /8,854 10⁻¹² 250 10⁻⁹ [- 7² + 11²] / 12²
E = 1.412 10⁴ N / C
6) As the field is positive, it is directed radially with direction coming out of the sphere