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nevsk [136]
3 years ago
11

We are constantly surrounded by many types of energy transfers and transformations, showing that energy in a system is conserved

. Which of these statements are true about how energy is conserved through transfers and transformations? Choose the two that apply. A. Energy is conserved when a lamp is turned on because some of the electrical energy is transferred into thermal energy. B. Energy is conserved when thermal energy is transferred from your body to a coat. C. Energy is conserved when some of the chemical energy from burning a candle is transformed into thermal energy, while the rest of the chemical energy is transformed into light energy. D. Energy is conserved when electrical energy from a hair dryer is transferred into thermal energy and then transformed by conduction.
Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

Energy is conserved when some of the chemical energy from burning a candle is transformed into thermal energy, while the rest of the chemical energy is transformed into light energy.

Energy is conserved when thermal energy is transferred from your body to a coat.

Explanation:

The principle of conservation of energy states that energy can neither be created nor destroyed, but can be transformed from one form to another. This can be regarded as the first law of thermodynamics.

In the answer, we see two major kinds of energy transformation around us. The both cases show the conservation of energy in a system.

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A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

6 0
2 years ago
a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin
creativ13 [48]
If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

                 (2.5 x 1 newton)  =  2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

                 (2.5 x 120 newtons)  =  300 newtons.

4 0
3 years ago
Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m
Leno4ka [110]

Answer: 1.593*10^{17} photons released if the laser is used 0.056 s during the surgery

                           

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

                            E =\frac{hc}{\lambda}

Where \lambda is the wavelength, h is the Planck's constant and  h is the speed of light

              h = 6.626*10^{-34} \frac{m^{2} kg }{s}  -> Planck's constant

              c = 3*10^{8} \frac{m}{s}  -> Speed of light

So, replacing in the equation:

                E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}

Then, the energy of each released photon by the laser is:

                E = 3.867*10^{-19} \frac{J}{photons}

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

               \frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

              2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}

And by doing a simple rule of three, if 2.844*10^{18} photons are released every second, then in 0.056 s:

            0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons are released during the surgery

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