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Contact [7]
3 years ago
7

How have increased carbon dioxide levels and temperatures affected living organisms?

Physics
2 answers:
balu736 [363]3 years ago
7 0
It leads to increase in temperature of the earth and it harms living being
Yuri [45]3 years ago
3 0

Answer:

it leads to increase in temperature of the earth and it harms living being

Explanation:

carbon dioxide is lite gas ,so it form a layer of carbon dioxide in the atmosphere.By our dairy activities such as driving ,burning etc release carbon dioxide which increase the layer of the carbon dioxide.the carbon dioxide layer absorb the heat energy from the sun so the temperature will increase and it will make global warming.the living organisms in the earth can not sustain the hot temperature hence it affect the living organisms in the earth

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3 years ago
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Siven an element's atomic number and mass number, how can you tell the number of protons and neutrons in its nucleus?
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3 years ago
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Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

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3 years ago
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Answer:

C. transverse

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3 years ago
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