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Gelneren [198K]
2 years ago
11

A lamp draws a current of 20 A when it is connected to a 100 V source? What is the resistance of the lamp?

Physics
1 answer:
S_A_V [24]2 years ago
7 0

Explanation:

R = V / I = 100V / 20A = 5 ohms.

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Infant car seats are made to face the rear of the car. This is safer in a front end collision because of Newton's First law. New
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Suppose a current flows through a copper wire. Which two things occur?
Bingel [31]

Answer:

a

Explanation:

as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it

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Direct current, DC, flows in a _________ direction while alternating current, AC, the direction of flow ___________. A) single,
kramer
The flow of Direct current (DC) is constant and flows in one direction. Most digital electronics make use of DC. Alternating current (AC) periodically flows in reverse and is mostly used to deliver power to houses, buildings and the like. With that alone, you can already rule out A, C and D. 

The answer would then be B. constant, periodically reversing. 
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A train of mass 5.6 × 10^5kg is at rest in a station.at time t=0s, a resultant force acts on the train and it starts to accelera
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Answer:

420000N

Explanation:

Given parameters:

Mass of the train  = 5.6 x 10⁵kg

Acceleration  = 0.75m/s²

Unknown:

Resultant force = ?

Solution:

According to newton's second law, force is the product of mass and acceleration;

   Force  = mass x acceleration

Resultant force that acts on the train is given below;

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7 0
3 years ago
The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra
Scrat [10]

Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

7 0
3 years ago
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