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3 years ago

What is the applied voltage of a circuit that has 10 amps of current and 12 ohms of resistance?

Physics

1 answer:

Anettt [7]3 years ago

7 0

Voltage (V)= Current (I) * Resistance (R)
V=IR=10*12=120volt

*This is a common answer of mine*
Hope this helps!

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A 50-kg copper block initially at 140Â°c is dropped into an insulated tank that contains 90 l of water at 10Â°c. Determine the f

xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

Temperature of block $T_b =140 \textdegree C$

Volume of water $V= 90L$

Temperature of water $T_w=10 \textdegree C$

Density of water $\rho=1000kg/m^3$

Specific heat of water $C_w=4.18KJ/kg-k$

Specific heat of copper $C_p=0.96KJ/kg-k$

Generally the equation for equilibrium stage is mathematically given by

$mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)$

$50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)$

$48(140-T_f)=376.2(T_f-10)$

$140-T_f=7.8375(T_f-10)$

$140-T_f=7.8375T_f-78.375$

$-8.8375T_f=-218.375$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=24.71$

6 0

3 years ago

Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the

True [87]

Answer:

$E=54V/cm$

$\Delta V=496.8V$ between the plates.

Explanation:

The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

$\Delta V=Ed$

So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

$E=\frac{\Delta V}{d}=\frac{475 V}{8.8cm}=54V/cm$

And we use the fact that the plates are 9.2cm apart to calculate the voltage between them:

$\Delta V=Ed=(54V/cm)(9.2cm)=496.8V$

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