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Akimi4 [234]
3 years ago
13

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

Physics
1 answer:
Mashcka [7]3 years ago
4 0

<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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Answer:

T = 2.4 + 2.4 = 4.8 [s]

Explanation:

In order to solve this problem, we must use the following kinematics equation and calculate the acceleration value.

x=x_{o} +v_{o}*t+(\frac{1}{2})*a*t^{2}

Vo = inital velocity = 0

x - xo = 15 [m]

t = time = 2.4 [s]

15 = 0.5*a*(2.4)^2

a = 5.208 [m/s^2]

We can use the same equation to find the time.

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t = 2.4 [s]

T = 2.4 + 2.4 = 4.8 [s]

7 0
3 years ago
n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
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Answer:

a). V = 3.13*10⁶ m/s

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q = +2e

M = 4.0u

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V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

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T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

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