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ASHA 777
3 months ago
15

g The current density in the 2.9-mm-diameter wire feeding an incandescent lightbulb is 0.33 MA/m2. Part A What's the current den

sity in the lightbulb's filament, whose diameter is 0.055 mm
Physics
2 answers:
Yuliya22 [3.3K]3 months ago
5 0

Answer:

j_{B} = 917.454\,\frac{mA}{m^{2}}

Explanation:

Current can be determined by multiplying current density by the cross-sectional area. The new current density can be derived through the subsequent relationship, while assuming that the same current flows through the new wire:

j_{A}\cdot A_{A} = j_{B}\cdot A_{B}

j_{B} = j_{A}\cdot \frac{A_{A}}{A_{B}}

j_{B} = j_{A} \cdot \left(\frac{D_{A}}{D_{B}} \right)^{2}

j_{B} = \left(0.33\,\frac{mA}{m^{2}} \right)\cdot \left(\frac{2.9\,mm}{0.055\,mm} \right)^{2}

j_{B} = 917.454\,\frac{mA}{m^{2}}

Ostrovityanka [3.2K]3 months ago
3 0

Answer:

Explanation:

Provided data:

Light bulb:

Diameter, d = 2.9 mm

Current density, J = 0.33 MA/m²

Filament:

Diameter, d = 0.055 mm

Current, I = J × A

Area = pi/4 × d²

= pi/4 × (0.0029)²

= 6.6 × 10^-6 m²

The current of the light bulb, I = 3.3 × 10^5 × 6.6 × 10^-6

= 2.18 A

Given that the current in the lightbulb equals the current in the filament, we find the filament current to be 2.18 A

Area = pi/4 × d²

= pi/4 × (5.5 × 10^-5)²

= 2.38 × 10^-9 m²

Current density, J = 2.18/2.38 × 10^-9

= 9.18 × 10^8 A/m²

= 918 MA/m²

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Keith_Richards [3271]
Consider the diagram below.

m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ =  acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted

Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t²                  (1)
x = 0.5a₂t²                  (2)

F = m₁a₁ = m₂a₂         (3)
Additionally,
x + y = 14000 m          (4)

From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²

From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s

Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m

The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.

Result: The starship shifts by 16.5 m (to the nearest tenth)

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The correct choice is B:)
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1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
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