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3 years ago

7

How does uniform motion relate to velocity and acceleration?

1 answer:

nekit [7.7K]3 years ago

8 0

-- Velocity is the speed and direction of motion.

-- Acceleration is any change of velocity ...
. . . speeding up, slowing down, or changing direction.

-- Uniform motion means no change ...
. . . straight line motion at constant speed
. . . constant velocity and zero acceleration.

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What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms

Mashcka [7]

<h2> $\bf{ \underline{Given:- }}$ </h2>

$\sf• \: The \: current \: in \: a \: circuit \: is \: 5 \: amps. \: and \: resistance \: is \: 30 \: Ohms.$

$\\$

<h2> $\bf{ \underline{To \: Find :- }}$ </h2>

$\sf{• \: The \: Potential \: Difference. }$

$\\$

$\huge\bf{ \underline{ Solution:- }}$

$\sf According \: to \: the \: question,$

$\sf• \: Current \: (I) = 5 \: Amps.$

$\sf• \: Resistance \: (R) = 30 \: Ω$

$\sf{Potential \: difference \: means \: Voltage \: ( V).}$

$\sf{We \: know \: that, }$

$\bf \red{ \bigstar{ \: V = IR }}$

$\rightarrow \sf V =5 \times 30$

$\rightarrow \sf V =150$

$\\$

$\sf \purple{Therefore, \: the \: potential \: difference \: is \: 150 \: v \: .}$

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2 years ago

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2 years ago

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3 years ago

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On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

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3 years ago