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3 years ago

Waste reduction + rationalization of consumption + reuse + recycling / ways of preserving natural resources

Physics

1 answer:

forsale [732]3 years ago

5 0

Answer:

Hello your question is vague hence I will provide a general answer on the importance of : Waste reduction, rationalization of consumption, reuse, recycling as ways of preserving the natural resources

answer :

Waste reduction : when we reduce the amount of wastage on items we use especially items produced from natural raw materials we will help preserve the natural resource because they can be used to produce varieties of other items

Reuse and recycling of waste products help keep our natural environment healthy thus preserving our natural resources.

Explanation:

Waste reduction, rationalization of consumption, reuse and recycling are all ways of preserving our natural resources

Reuse and recycling of waste products help keep our natural environment healthy thus preserving our natural resources.

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Why must objects be cooled before their mass is determined on a sensitive balance?

zubka84 [21]

Objects should be cooled before their mass is determined on a sensitive balance because it could damage the balance. Also, because it would give you wrong reading of the mass. Hot objects would warm the air around it. A warm air would expand and would produce convection as it rises causing to give the object a mass that is less than the actual. Another reason would be it would cause instability in the readings, the mass would fluctuate every now and then due to the convection currents around the object. It is always recommended to weigh the masses of objects that are in room temperature.

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3 years ago

Hi free ponits hehehehhehbhrgivudksjbtyuvwijfe

Vilka [71]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

1. Infer how the height of the lines on a seismograph change with an increase in

SOVA2 [1]

Answer:

his movement is proportional to the intensity of the earthquake,

Explanation:

An earthquake is a record of the intensity of an earthquake as a function of time.

Where the intensity is plotted on the y-axis, which corresponds to the vertical movement of the detector, this movement is proportional to the intensity of the earthquake, therefore the intensity increases the amplitude of the oscillation increases.

And the in x corresponds to time

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3 years ago

Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What i

OverLord2011 [107]

Answer: $2.89(10)^{-3} m$

Explanation:

The Heisenberg uncertainty principle postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.

In other words:

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

$\Delta x \geq \frac{h}{4 \pi m \Delta V}$ (1)

Where:

$\Delta x$ is the uncertainty in the position of the electron

$h=6.626(10)^{-34}J.s$ is the Planck constant

$m=9.11(10)^{-31}kg$ is the mass of the electron

$\Delta V$ is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is $0.001\%$ of the velocity of the electron $V=2 km/s=2000 m/s$ , then $\Delta V$ is:

$\Delta V=2000 m/s(0.001\%)$

$\Delta V=2000 m/s(\frac{0.001}{100})$

$\Delta V=2(10)^{-2} m/s$ (2)

Now, the least possible uncertainty in position $\Delta x_{min}$ is:

$\Delta x_{min}=\frac{h}{4 \pi m \Delta V}$ (3)

$\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)}$ (4)

Finally:

$\Delta x_{min}=2.89(10)^{-3} m$

5 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

Time taken by the rocket to fall from maximum height is 20.29 seconds

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago