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3 years ago

Which fossil fuel is produced as a by-product that occurs when bacteria decompose organic material under anaerobic conditions?

1 answer:

5 0

Bio-gas is the naturally produced fossil fuel, a by-product when bacteria decompose organic material under anaerobic conditions.

<h3><u>Explanation:</u></h3>

Organic matter particularly waste material is broken down by bacteria through fermentation in an environmental condition without any presence of oxygen. This process of decomposition leads to formation of bio-gas with "carbon dioxide and methane" in a 2:3 ratio.

The above biological process is termed as bio-digestion or anaerobic digestion. Methane is flammable and thus bio-gas can be used as "energy source", a waste-to-energy transformation. The remaining decomposed matter is ideal as manure for plants due to its rich nutrient level.

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At this output power, how long would it take an 89 kg cyclist (including the mass of his bike) to climb the famed 1100-m-high al

Artemon [7]

<span>In general the route that follows the 1100-m-high alpe d'huez mountain stage takes anywhere from four to six hours in athletes from around the world. A beginner would probably take 8 hours or so. World records were recently broken with a time just under three hours. The most friction for cyclist is the competition itself.</span>

7 0

3 years ago

An elevator cab has a mass of 4710 kg and can carry a maximum load of 1910 kg. If the cab is moving upward at full load at 3.80

OLEGan [10]

Answer:

Power P=246.78 KW

Explanation:

Mass of Elevator=4710 Kg

Can carry load =1910 Kg

Total Mass m to be lifted =4710 kg + 1910 Kg =6620 kg

speed = 3.80 m/s ⇒ per second distance traveled (height h) = 3.80 m and time t = 1 s

Using formula of Power P = $\frac{work}{time}$

P = $\frac{mgh}{t}$

P= 6620kg × 9.81 m/s² × 3.80 m / 1s

P= 246780.36 W

P=246.78 KW

4 0

3 years ago

You don;t run across elastic collision in real life

pentagon [3]

Answer:

no comment sorry bro

Explanation:

4 0

3 years ago

PHYSICS PLEASE HELP 80 PTS!!!!!!

VladimirAG [237]

1. B) 32 m/s

The speed of the baseball is given by

$v=\frac{d}{t}$

where

d = 48 m is the distance travelled

t = 1.5 s is the time taken

Substituting,

$v=\frac{48}{1.5}=32 m/s$

2. C) 12 m/s

The average velocity is given by

$v=\frac{d}{t}$

where

d = 60 - 0 = 60 m is the displacement

t = 5 s is the time interval

Substituting,

$v=\frac{60}{5}=12 m/s$

3. Missing diagram

4. D) velocity

In fact, velocity is defined as the ratio between the displacement ( $\Delta s$ ) and the time interval ( $\Delta t$ ) needed to achieve that change in position:

$v=\frac{\Delta s}{\Delta t}$

Similarly, acceleration is defined as the ratio between the change in velocity ( $\Delta v$ ) and the time interval ( $\Delta t$ ):

$a=\frac{\Delta v}{\Delta t}$

Comparing the two equations, we see that displacement is to velocity as velocity is to acceleration.

5. C) 5/1 cm/year

We know that the ridge moves 25 cm in 5 years, so we have:

- Displacement (d): d = 25 cm

- Time interval (t): t = 5 y

Using the equation for the velocity:

$v=\frac{d}{t}=\frac{25 cm}{5 y}=5 cm/y$

6. D

The missing graph is attached. Each graph represents a distance (on the y-axis, measured in metres) versus a time (on the x-axis, measured in seconds). Therefore, the speed in each graph can be simply calculated from the slope of the curve, since speed is the ratio between distance and time:

$v=\frac{\Delta y}{\Delta x}$

For each graph:

A. $\frac{\Delta y}{\Delta x}=\frac{8.6-2}{10}\sim 0.66$

B. $\frac{\Delta y}{\Delta x}=\frac{5.7-0}{10}\sim 0.57$

C. $\frac{\Delta y}{\Delta x}=\frac{9.6-3}{10}\sim 0.66$

D. $\frac{\Delta y}{\Delta x}=\frac{7.5-0}{5}\sim 1.5$

So we can say that the correct graph must be graph D, the closest to 1.57.

7. B) The car has come to a stop and has zero velocity

The missing graph is attached.

The graph shows the position of the car versus time. From the graph, we can see that in segment B, the position of the car does not change: this means that its velocity is zero, since the velocity is defined as the ratio between the change in position (displacement) and the time interval:

$v=\frac{\Delta s}{\Delta t}$

And since the displacement is zero, the velocity is also zero.

8. C) from 4.5 to 5.0 seconds

Missing graph is attached.

The graph represents the distance covered versus the time taken: therefore, the speed of the ball can be evaluated from the slope of the curve.

The only region in which the slope of the curve is zero is between 4.5 seconds and 5.0 seconds: therefore, this is the region where the soccer ball is not moving.

9. D) diagonal line with varying slope, from 3 to 4.

The table of data is:

Time (s) 0 1 2 3 4

Distance (m) 0 3 6 12 16

We want to know how the distance/time graph would like like. We have:

- At t = 1, the distance is $d=3$ , so the slope is $\frac{d}{t}=\frac{3}{1}=3$

- Similarly, at t = 2: $\frac{d}{t}=\frac{6}{2}=3$

- Instead, at t =3, the slope is $\frac{d}{t}=\frac{12}{3}=4$

- Similarly, at t=4, $\frac{d}{t}=\frac{16}{4}=4$

So the correct answer is D.

10. D) To move in a circle requires an acceleration and therefore a net force. This force is supplied by the slingshot.

For an object in circular motion, the velocity constantly changes (because the direction changes): this means that an acceleration is required (towards the centre of the circle), and therefore a force must be exerted (also towards the centre of the circle) to keep the object in the circular path.

As soon as this force is removed, the object is "free" to leave the circular trajectory and continue its motion following a straight path at constant velocity, according to Newton's first law.

11. B) The acceleration will become 1/4 as much.

Re-arranging Newton's second law formula,

$F=ma \rightarrow a = \frac{F}{m}$ (1)

The force exerted on the satellite is actually the force of gravity:

$F=\frac{Gm M}{d^2}$ (2)

Substituting (2) into (1),

$a=\frac{F}{m}=\frac{GM}{d^2}$

We see that there is no dependance on the mass of the satellite (m), but only on the distance. Since in this problem the distance is doubled (d'=2d), the new acceleration will be:

$a'=\frac{GM}{(2d)^2}=\frac{1}{4}(\frac{GM}{d^2})=\frac{a}{4}$