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3 years ago

Agricultural waste used as a fuel is an example of

2 answers:

6 0

<span>Agricultural waste used as a fuel is an example of biofuels, there already different kinds of biofuels that came from agricultural waste, the most common are the biogas and bioethanol. biogas are produce when agricultural waste such as poultry manures and grassy waste are fermented to create biogas. while biothanol uses sugar rich agricultural waste to make ethanol</span>

jeka57 [31]3 years ago

3 0

biofuels is the answer to your question

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A 76 N crate is hung from a spring

lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0

3 years ago

The impulse experienced by a body is equivalent to the body’s change in

nika2105 [10]

<h2>Answer: </h2>

Momentum

<h2>Explanation: </h2>

The momentum of a particle is defined as the product of the particle mass and the particle velocity as follows:

$\overrightarrow{p}=m\overrightarrow{v}$

On the other hand, the impulse of a constant force is defined as:

$\overrightarrow{J}=\varSigma\overrightarrow{F}(t_{2}-t_{1})=\varSigma\overrightarrow{F}\Delta t$

We also know that the net force acting on a particle equals the rate of change of the particle’s momentum, so:

$\varSigma\overrightarrow{F}=m\overrightarrow{a}=m\frac{d}{dt}(\overrightarrow{v})=\frac{d}{dt}(m\overrightarrow{v})=\frac{d\overrightarrow{p}}{dt}$

If the force is constant, then $\frac{d\overrightarrow{p}}{dt}$ equals the total change in momentum over a period of time:

$\varSigma\overrightarrow{F}=\frac{\overrightarrow{p_{2}}-\overrightarrow{p_{1}}}{t_{2}-t_{1}} \\ \\ \varSigma\overrightarrow{F}(t_{2}-t_{1})=\overrightarrow{p_{2}}-\overrightarrow{p_{1}} \\ \\ \boxed{\overrightarrow{J}=\Delta \overrightarrow{p}}$

3 0

3 years ago

Design an inverting amplifier that has a gain of -47 (this gain is negative). Pick resistor values that you have in the lab kit.

Nostrana [21]

Answer:

R2= 470 ohms R1= 10ohms. Schematic is attached

Explanation:

Av=-(R2/R1)

-47=-(470/10)

-47=-47

7 0

3 years ago

when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo

Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

$F = k\frac{|q_1|.|q_2|}{r^2}$

Where,

k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

$\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1$ ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

$q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}$

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

$\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6$ .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

$-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123$

$q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\$

6 0

4 years ago

How far back into the universe are we able to see? Explain.

Nadusha1986 [10]

Answer:

About 46 billion light years Because space expands so what we didn’t see before is what we could see now. An example could be how everyone thought Pluto was a regular sized planet but once you get a look at the bigger picture and compare it to other thing in space we now know Pluto is a dwarf planet.

3 0

3 years ago