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3 years ago

Nonrenewable resources are natural resources that are limited in supply and cannot be replaced as quickly as they are used up.

Physics

1 answer:

zysi [14]3 years ago

8 0

Answer:

True

Explanation:

Nonrenewable resources ARE limited in supply. They don't get replaced at the speed they get made. For example: we pump crude oil from the ground at a rate that makes it impossible for crude oil to be replaced. Crude oil takes millions of years to produce

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) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe

pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0

3 years ago

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(25 pts) Estimate how much collector area and storage capacity would be required for an active solar hot-water system designed t

Aneli [31]

Answer:

The required total area is 1.48 m²

Explanation:

Given that,

Latitude = 44+° N

New Mexico,

Latitude= 35+° N

Heat capacity = 4200 J/Kg°C

Temperature = 60°C

Let us assume the input temperature 22°C

Estimate volume of water 100 ltr for 4 person.

We need to calculate the heat

Using formula of heat

$H=mc_{p}\Delta T$

$H=mc_{p}(T_{f}-T_{i})$

Put the value into the formula

$H=100\times4200\times(60-22)$

$H=15960\ KJ$ ...(I)

Let solar radiation for 6 hours/day.

We need to calculate the total energy per unit area

Using formula of energy

$E=1000\times6\times3600\ J/m^2$

$E=21600\ KJ/m^2$

Let the efficiency of collector is 50 %

Then, the total energy per unit area will be

$E=21600\times\dfrac{50}{100}$

$E=10800\ KJ/m^2$ ....(II)

We need to calculate the required total area

Using equation (I) and (II)

$A=\dfrac{H}{E}$

Where, H = heat

E = total energy

Put the value into the formula

$A=\dfrac{15960}{10800}$

$A=1.48\ m^2$

Hence, The required total area is 1.48 m²

6 0

3 years ago

what is cut off from the open sea coral reefs or sandbars A watershed B lagoon C reef or D atoll . also this is a science questi

Klio2033 [76]

Hello there.

Its D.

HOPE I HELPED

7 0

3 years ago

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A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

Drug abuse is characterized by _____.

arlik [135]

There can be mental health effects and psychological dependence

8 0

3 years ago

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