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DaniilM [7]
2 years ago
10

16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current d

rawn b) The quantity of electricity in Coulombs taken in 5 minutes c) The energy absorbed in 100 hours.
Physics
1 answer:
dsp732 years ago
5 0

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t    = power X time

W / t = P   power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec  = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c)  E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

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An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is
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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

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\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

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\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

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\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

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\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

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\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

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\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

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\dashrightarrow\:\: \sf{m = -2}

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<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

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3 0
3 years ago
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Answer:0.114 C

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Let q_1 and q_2 be the charges such that

q_1+q_2=4.7

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F=\frac{kq_1q_2}{r^2}

4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}

q_1\cdot q_2=0.522

put the value of q_1

q_2\left ( 4.7-q_2\right )=0.522

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3 0
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Answer:

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4 0
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