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3 years ago

The production of carbon dioxide might be an indicator of what process

Physics

2 answers:

disa [49]3 years ago

6 0

Its a proces of combustion "burning"humans produce carbon dioxide,rocks<plants whatever is burning energy

aalyn [17]3 years ago

3 0

The carbon dioxide is produced by two methods naturally, which is either photosynthesis or by decomposition. The indicator thus for the carbon dioxide production be the decomposition most likely.

<u>Explanation:</u>

The carbon dioxide is produced by the process of decomposition and photosynthesis both. But in the process of photosynthesis it is released as a by product which then consumed in the process of respiration. Therefore if there is ample production, it indicated of decomposition. The decomposition is the decaying process of organic matter.

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1) The current in a light bulb is 0.335 Amps. How long does it take for a total charge of 2.76 C to

uysha [10]

Answer:

Time=8.23880597 seconds

Explanation:

Quantity of charge(q)=2.76c

Current(I)=0.335A

Time(t)=?

t=q/I

t=2.76/0.335

t=8.23880597seconds

7 0

3 years ago

How to find new pressure of gas if bottle explodes

KIM [24]

If the container explodes there is no pressure, becuase all your gas has escaped its container, there for, you ain’t got no gas

6 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

which of the following is the correct si unit to use in measuring the mass of a boulder a. tons b. kilograms c. ounces d. pounds

Salsk061 [2.6K]

B. kilograms iuyiuiuhikj

7 0

3 years ago

Read 2 more answers

In odd nuclei, what determines the final spin of the nucleus?

Katen [24]

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

$1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }$

Here,

$j=\frac{1}{2}$

and, L=1.

Fort parity,

$(-1)^{L}$

Put the value of L.

Parity will be -1.

Now, spin will be

$S=(\frac{1}{2} )^{-1}$ .

7 0

3 years ago