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Marta_Voda [28]
3 years ago
11

hich one of the following statements could be an operational definition of electric current?View Available Hint(s)Which one of t

he following statements could be an operational definition of electric current?The magnitude of the physical quantity of electric current I in a wire equals the magnitude of the electric charge q that passes through a cross section of the wire divided by the time interval Δt needed for that charge to pass.
Physics
1 answer:
klemol [59]3 years ago
8 0

Answer:

it is True as the operational definition of electric current.

Explanation:

The definition of electric current is

         I = dQ / dt

By convention the direction of the current is the direction in which a positive charge flows.

The initial expression is the derivative that is the change of the load in the unit of time and this occurs in a given cross-sectional cable.

The proposed definition is the same as this, so it is True as the operational definition of electric current.

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Your answer:

150 kilometer's per hour.

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A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
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To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

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3 years ago
A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward al
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Answer: option 1 : the electric potential will decrease with an increase in y

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V = kq/y

Where k = 1/4πε0

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From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

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