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3 years ago

What is the equation for calculating the heat energy released during heating effect of electric current?

2 answers:

5 0

The heat energy released from a piece of wire or any other section of a circuit is:

Energy = (voltage between its ends) x (current through it) x (time it's been going)

anygoal [31]3 years ago

3 0

The equation for calculating the heat energy released by the heating effect of electric current is given by the following equation : -

H = I²Rt

Where H stands for heat energy, I for electric current, R for resistance and t for time.

You might be interested in

Amber is a (a) metal (b) rubber (c) resin (d) sugar

zlopas [31]

Answer:

c. Resin.

Explanation:

Amber is a resin.

4 0

2 years ago

A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average

Maksim231197 [3]

Answer: $72.66 AU=1.089(10)^{10} km$

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period $T$ of a planet is proportional to the cube of the semi-major axis $a$ of its orbit”:

$T^{2}\propto a^{3}$ (1)

Now, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes:

$T^{2}=a^{3}$ (2)

So, knowing $T=619.36 years$ and isolating $a$ from (2) we have:

$a=\sqrt[3]{T^{2}}$ (3)

$a=\sqrt[3]{(619.36 years)^{2}}$ (4)

Finally:

$a=72.66 AU$ T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

$a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km$

4 0

3 years ago

A 10.0 N package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is su

inna [77]

Answer:

0.025 m

Explanation:

From the question,

Applying Hook's law

F = ke................... Equation 1

Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.

make e the subject of the equation

e = F/k....................... Equation 2

Given: F = 10 N, e = 395 N/m

Substitute these values into equation 2

e = 10/395

e = 0.025 m

7 0

3 years ago

A distorted 1kHz sine wave is represented by the equation: LaTeX: v(t) = 5sin(2 \pi 1000t) + 0.05sin(2 \pi 3000t)v ( t ) = 5 s i

vovikov84 [41]

Answer:

As given in the problem statement

frequency=1 KHz=1*10^3 Hz

V(t) is represented as

v(t) = 5sin(2 \pi 1000t) + 0.05sin(2 \pi 3000t)

v ( t ) = 5 s i n ( 2 π 1000 t ) + 0.05 s i n ( 2 π 3000 t )

Total harmonic distortion will be 234 Pi

3 0

2 years ago

A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago