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3 years ago

What are the strengths and weaknesses of the four methods of waste management?

1 answer:

4 0

Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;

* It creates employment

* It keeps the environment clean

* The practice is highly lucrative

* It saves the earth and conserves energy

The weaknesses of the methods of waste management includes;

* The sites are often dangerous

* The process is mostly

* There is a need for global buy-in

* The resultant product had a short life

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Slope or velocit ime graph represent

Lubov Fominskaja [6]

Answer:

Please check the explanation.

Explanation:

The slope of the velocity-time graph illustrates the change in velocity with respect to change in time.

In other words, the acceleration of the object is defined by the slope of a velocity graph. The acceleration can be obtained by finding the slope at a particular time.

Hence, the slope of the velocity time graph represent represents acceleration.

2. Also the acceleration of any object, like a car or bike, at some discrete instant in time 't' is termed as instantaneous acceleration which can be determined if we may take the derivate of the given velocity function.

3. A vector that has a direction and magnitude of 1 is termed as a unit vector, often called a direction vector.

7 0

3 years ago

A piston-cylinder device initially contains 0.3 kg of nitrogen gas at 160 kPa and 140oC. The nitrogen is now expanded isothermal

irakobra [83]

Answer:

<h2>Work done by the gas is given as</h2><h2> $W = 1.72 \times 10^4 J$ </h2>

Explanation:

As we know that the process is isothermal so here work done is given as

$W = nRTln(\frac{P_1}{P_2})$

here we know that

$n = \frac{w}{M}$

$n = \frac{300}{28} = 10.7 moles$

now we have

$W = 10.7 (8.31)(273 + 140) ln(\frac{160}{100})$

so we have

$W = 1.72 \times 10^4 J$

4 0

3 years ago

When a lorge heavy truck collieds with a

arsen [322]

Yes since the heavy truck it’s way bigger it can cause more damage to the passenger car

7 0

3 years ago

1 point

katrin [286]

Answer:

Phosphorus

Explanation:

Phosphorus needs 3 electrons to complete its outter most shell of electrons

8 0

3 years ago

To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad

Julli [10]

The magnitude of the forces acting at the top are;

$\mathbf{F_{Top, \ x}}$ = 132.95 N

$\mathbf{F_{Top, \ y}}$ = 0

The magnitude of the forces acting at the bottom are;

$\mathbf{F_{Bottom, \ x}}$ = $\mathbf{ F_f}$ = -132.95 N

$\mathbf{F_{Bottom, \ y}}$ = 784.8 N

The known parameters in the question are;

The mass of the person, m₁ = 70.0 kg

The length of the ladder, l = 6.00 m

The mass of the ladder, m₂ = 10.0 kg

The distance of the base of the ladder from the house, d = 2.00 m

The point on the roof the ladder rests = A frictionless plastic rain gutter

The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder

The location of the point the person is standing = 3 meters from the bottom

g = The acceleration due to gravity ≈ 9.81 m/s²

The required parameters are;

The magnitudes of the forces on the ladder at the top and bottom

The strategy to be used;

Find the angle of inclination of the ladder, θ

At equilibrium, the sum of the moments about a point is zero

The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C

Taking moment about the point of contact of the ladder with the ground, <em>B </em>gives;

$\sum M_B$ = 0

Therefore;

$\sum M_{BCW}$ = $\sum M_{BCCW}$

Where;

$\sum M_{BCW}$ = The sum of clockwise moments about <em>B</em>

$\sum M_{BCCW}$ = The sum of counterclockwise moments about <em>B</em>

Therefore, we have;

$\sum M_{BCW}$ = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

$\sum M_{BCCW}$ = $F_R$ × √(6² - 2²)

Therefore, we get;

2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = $F_R$ × √(6² - 2²)

$F_R$ = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95

The reaction force on the wall, $F_R$ ≈ 132.95 N

We note that the magnitude of the reaction force at the roof, $F_R$ = The magnitude of the frictional force of bottom of the ladder on the floor, $F_f$ but opposite in direction