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igor_vitrenko [27]
3 years ago
12

At room temperature what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm that is needed to cau

se a 2.75 a current to flow?
Physics
1 answer:
Dima020 [189]3 years ago
6 0
Electric field strength = resistivity of copper x current density
where
p= 1.72 x 10^-8 <span>ohm meter
diameter = 2.05mm=.00205 m
current = 2.75 A
</span>get first the current density:
current density = current/ cross section area
find the cross section area
cross section area = pi.(d/2)^2;  
cross section = 3.3 006x10-6 m^2
substitute the values 
current density = 2.75A/3.3006x 10-6m^2
current density=35.55 x1 0^2 A/m^2
Electric field stregnth =1.72 x 10^-8 ohm meter x 35.55 x10^2 A/m^2
Electric field stregnth= 46.415 Volts/m

The electric field strength of copper is 46.415 V/m.


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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
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ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

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Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

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A train increases its speed steadily from 10 m/s to 20 m/s in
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Answer:

15m/s

Explanation:

add the two speeds and divide by 2

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30/2=15

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