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2 years ago

A substance burns in the presence of oxygen.

Physics

2 answers:

valina [46]2 years ago

5 0

If I don’t know a answer I put C bc it is a 50/50 chance but it might be wrong

makkiz [27]2 years ago

5 0

Answer:

D. flammability

Explanation:

I took the test and got it right

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Why can biofuels be described as renewable resources?

gulaghasi [49]

Answer:

Biofuel is fuel that is produced through contemporary processes from biomass, rather than by the very slow geological processes involved in the formation of fossil fuels, such as oil. Since biomass technically can be used as a fuel directly, some people use the terms biomass and biofuel interchangeably.

Biofuel, any fuel that is derived from biomass—that is, plant or algae material or animal waste. Since such feedstock material can be replenished readily, biofuel is considered to be a source of renewable energy, unlike fossil fuels such as petroleum, coal, and natural gas.

stream life goes on and dynamite

we're about to reach 1B

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5 0

2 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

2 years ago

Cobaltâ’60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this

Crank

Energy of gamma rays is given by equation

$E = h\nu$

here we know that

h = Planck's constant

$\nu = frequency$

now energy is given as

$E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}$

$E = 7.52 \times 10^{-13} J$

now by above equation

$E = h\nu$

$7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu$

$\nu = 1.14 \times 10^{21} Hz$

now for wavelength we can say

$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}$

$\lambda = 2.63 \times 10^{-13} m$

3 0

2 years ago

You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through

Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

$V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA$

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0

2 years ago

The A note has a frequency of 440 Hz. What is the wavelength if the speed of sound is 343 m/s?

Nina [5.8K]

The wavelength of a wave (λ) is given by λ $= \frac{c}{f}$
where c is the wave speed and f is the frequency

7 0

2 years ago

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