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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.

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A force acting on a body causes a change in the momentum of the from 12 kgms-1 to 16 kgms-1 in 0.2 s. calculate the magnitude of

Nonamiya [84]

Answer: Impulse = 4 kgm/s

Explanation:

From the question, you're given the following parameters:

Momentum P1 = 12 kgm/s

Momentum P2 = 16 kgm/s

Time t = 0.2 s

According to second law of motion,

Force F = change in momentum ÷ time

That is

F = (P2 - P1)/t

Cross multiply

Ft = P2 - P1

Where Ft = impulse

Substitute P1 and P2 into the formula

Impulse = 16 - 12 = 4 kgm/s

The magnitude of the impulse is therefore 4 kgm/s.

6 0

3 years ago

What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two

lesya [120]

Answer:

Power = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

$\beta (dB)= 10 \ log_{10 } (\dfrac{I}{I_o})$

where;

the threshold of hearing $I_o = 10^{-12} (W/m^2)$

$\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})$

$10^{10.2} = \dfrac{I}{10^{-12}}$

$I = 10^{10.2} \times 10^{-12}$

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power = 124.50 W

7 0

3 years ago

A piece of copper metal is initially at 100.0°C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a tem

olya-2409 [2.1K]

Answer:

800j/k

Explanation:

6 0

3 years ago

Read 2 more answers

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

2 years ago

A car slams on its breaks,producing friction between the tires and the road.Into which type of energy is the mechanical energy o

Step2247 [10]

The answer is Heat Energy

4 0

2 years ago

Read 2 more answers