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2 years ago

Which of these would not contribute to a global increase in temperature?

Physics

1 answer:

Katyanochek1 [597]2 years ago

6 0

Answer:

planting trees - increasing the amount of animal life on earth

Explanation:

from all the other options given, all except planting trees will increase global temperature.

1. deforestation: this removes vegetative cover and the amount of oxygen in the atmosphere.

2. increasing the use of gasoline burning vehicles: burning gasoline produces carbon dioxide which is a green house gas which traps heat in the atmosphere and causes global warming

3. increasing the industrial release of carbon dioxide: carbon dioxide is released when gasoline is burned in industries, this in-turn causes global warming.

Trees produces oxygen and sucks in carbon dioxide which causes global warming. more trees mean more oxygen and less carbon dioxide which translates to less global increase in temperature

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Two small children decide it would be fun to toss a couple of large cats at each other. Cat A (7kg) is thrown at 7m/s and cat B

Alex777 [14]

Answer:

V=4.7m/s

Explanations:

Let Ma mass of cat A=7kg

Va velocity of cat A=7m/s

Mb mass of cat b=6.1kg

VB velocity of cat b=2m/s

From conservation of linear momentum

MaVa+MbVb=(Ma+Mb)V

7*7+6.1*2=(7+6.1)V

61.2=13.1V

V=4.7m/s

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2 years ago

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.

Thus,

r = $\frac{mv}{qB}$ ÷ sinΘ

But, sinΘ = sin $90^{0}$ = 1.

So that;

r = $\frac{mv}{qB}$

= (9.11 × $10^{-31}$ × 121) ÷ (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ )

= 1.10231 × $10^{-28}$ ÷ 2.608 × $10^{-24}$

= 4.2266 × $10^{-5}$

= 4.23 × $10^{-5}$ m

The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

B. The frequency 'f' of the motion is called cyclotron frequency;

f = $\frac{qB}{2\pi m}$

= (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ ) ÷ (2 × $\frac{22}{7}$ × 9.11 × $10^{-31}$ )

= 2.608 × $10^{-24}$ ÷ 5.7263 × $10^{-30}$

= 455442.4323

f = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0

3 years ago

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2 years ago

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Alenkinab [10]

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