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jek_recluse [69]
3 years ago
10

How often is carbon dioxide cycled through the atmosphere

Physics
1 answer:
olga55 [171]3 years ago
5 0

Maybe around 350 years, depending on the carbon cycle and the time taken through steps.

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If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means
lilavasa [31]

Answer:

M V = m v        conservation of momentum (Caps-cannon  Small-projectile)

V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s    recoil velocity of cannon

KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2  = 40 kg m^2/s^2 = 40 J

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3 years ago
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A child in danger of drowning in a river is being carried downstream by a current that flows due south uniformly with a speed of
tia_tia [17]

Let the rescue boat starts at an angle theta with the North

now its velocity towards East is given as

v_x = 24sin\theta

v_y = -24cos\theta + 3

now in some time "t" it will catch the boy

so we will have

t = \frac{0.5}{24sin\theta}

also we have

t = \frac{2}{-24cos\theta + 3}

now we have

\frac{2}{-24cos\theta + 3} = \frac{0.5}{24sin\theta}

4*24sin\theta = - 24cos\theta + 3

96 sin\theta + 24cos\theta = 3

by solving above we got

\theta = 164 degree

3 0
3 years ago
What is peer review?...
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3 years ago
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Michael is playing with two horseshoe magnets. He is trying to get them to touch, but they will not regardless of how hard he tr
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Since like poles repel, the two horseshoe magnets have like poles facing each other, hence they repel each other and therefore they will not come in contact
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If the car’s speed decreases at a constant rate from 71 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, ass
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Answer:

The acceleration and the distance are 25200 mi/h² and 0.1008 mi.

Explanation:

Given that,

Initial speed = 71 mi/h

Final speed = 50 mi/h

Time = 3.0 s

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Using equation of motion

v=u+at

a=\dfrac{v-u}{t}

Put the value in the equation

a=\dfrac{(50-71)\times3600}{3}

a=-25200\ mi/h^2

Negative sign shows the deceleration.

(b). We need to calculate the distance

Using equation of motion

v^2=u^2+2as

(50)^2=(71)^2+2\times(-25200)\times s

s=\dfrac{(50)^2-(71)^2}{-25200}

s=0.1008\ mi

Hence, The acceleration and the distance are 25200 mi/h² and 0.1008 mi.

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