A runner increases her speed from 8.9 m/s to 18.3 m/s in 8 seconds. What is the acceleration of the runner

Explain how carbon is cycled between the hydrosphere and geosphere. Use specific examples.

joja [24]

Carbon is found in the solid form in geosphere of our earth. Coal and oil are some of the examples of materials containing carbon in the geosphere. when the coal or oil is burnt, carbon dioxide is formed and released in atmosphere. This carbon dioxide is absorbed by the water of the hydrosphere with the help of algae and plankton. The water turns acidic in nature. This way carbon is transferred from geosphere to hydrosphere.

5 0

3 years ago

A nova occurs when

eimsori [14]

C. hydrogen accreted onto a white dwarf from a close companion rapidly fuses to helium, releasing a large amount of energy.

The accreted material, composed mainly of hydrogen, is compacted on the surface of the white dwarf due to the intense gravitational force on that place. As material accumulates, The white dwarf becomes increasingly hot, until it reaches the critical temperature for ignition of nuclear fusion.

5 0

3 years ago

7) A crazy cat (yes, this is redundant) is running along the roof of a 60 m tall building. The cat is moving at a constant veloc

gregori [183]

Answer:

The distance from the base of the building to the landing site is 154 m.

The total flight time is 3.5 s.

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).

v = velocity vector of the cat at time t.

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

r1y = y0 + v0y · t + 1/2 · g · t²

-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

- 60 m = -4.9 m/s² · t²

-60 m / - 4.9 m/s² = t²

t = 3.5 s

The cat reaches the ground in 3.5 s

Now, we can calculate the horizontal component of r1:

r1x = x0 + v0 · t

r1x = 0 m + 44 m/s · 3.5 s

r1x = 154 m

The distance from the base of the building to the landing site is 154 m.

The total flight time was already calculated and is 3.5 s.