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3 years ago

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when energy from battery was added to water, were the gases produced by this made of the same particles as those produced from h

eating the water?

1 answer:

Ilya [14]3 years ago

6 0

Batteries often convert chemical energy directly to electrical energy.

From the experiment, the constructed conclusion was that;

The Two different gases with different properties are made from adding energy from a battery to water.

The particles that make up these different gases is different from each other. They are different than the ones that were made from heating water.

The matter that is composed of substances have to originate from matter that made up some of the original water particles.

When energy from a battery is added to water, two streams of gas bubbles is said to be produced. Then a lit match is put into trapped gas from these two sources, one pops and the other glows brighter.

The study was to test the flammability of gases produced by providing energy to water with electricity.

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Studying this brochure from nasa, which explains more detail the instruments carried by the Juno spacecraft which scientific act

adoni [48]

Answer:

Juno scientific payload includes:

A gravity/radio science system (Gravity Science)
A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
A vector magnetometer (MAG)
Plasma and energetic particle detectors (JADE and JEDI)
A radio/plasma wave experiment (Waves)
An ultraviolet imager/spectrometer (UVS)
An infrared imager/spectrometer (JIRAM)

Explanation:

Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.

3 0

2 years ago

The typical unit for a period used with Kepler's third law is

ollegr [7]

Well, if you're using the law to work with periods of Earth satellites,
then the most convenient unit is going to be 'hours' for the largest
orbits, or 'minutes' for the LEOs.

But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.

6 0

3 years ago

Psuedopod in a sentence

mixas84 [53]

Does it have to be that exact word. cause it is just another term for psuedopodium

5 0

3 years ago

A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.

ANTONII [103]

<u>Answer:</u> The number of photons are $3.7\times 10^8$

<u>Explanation:</u>

We are given:

Wavelength of microwave = $1.22\times 10^8nm=0.122m$ (Conversion factor: $1m=10^9nm$ )

To calculate the energy of one photon, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = 0.122 m

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

4 0

3 years ago

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$

Solving for A

A = $\frac{H * L }{k * [ T_{H}- T_{C} ] }$

A = $\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}$

A = $\frac{85}{12550}$ = 6.77 × $10^{-3}$ m²

Now Area of cylinder is :

A = $\frac{\pi }{4}$ d²

solving for d:

d = $\sqrt{\frac{4 * 0.00677 }{\pi } }$

d = 9.28 cm

5 0

3 years ago