A 60 kg person sits in a chair. How much does the earth pull on the person? (Acceleration due to gravity is -10m/s2) F = mxa → F
1 answer:
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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
Answer:
, assuming that the gravitational field strength is
.
Explanation:
Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.
By Newton's Second Law, the net force on this block would be . External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.
Let denote the mass of this block. It is given that
. The weight of this block would be:
.
Hence, the force that the boy applies on this block would be upward with a magnitude of .
The mechanical work that a force did is equal to the product of:
- the magnitude of the force, and
- the displacement of the object in the direction of the force.
The displacement of this block (upward by ) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:
- the magnitude of the force that this boy exerted,
, and
- the displacement of this block in the direction,
.
.