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3 years ago

Reducing waste and recycling things are ways to conserve energy. True false

Physics

2 answers:

Leya [2.2K]3 years ago

6 0

I think the correct answer is true :)

nydimaria [60]3 years ago

5 0

True b/c we conserve energy by not using energy to make new materials because we have those available because we recycled

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Alexeev081 [22]

I included things
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7 0

2 years ago

Which of the following supports the theory of continental drift?

Y_Kistochka [10]

I think it's The fossil record. The same animal fossil is in Africa and South America. The animal could have not swim across so its the fossil record

7 0

2 years ago

Electricity costs 6 cents per kilowatt hour. In one month one home uses one megawatt hour of electricity. How much will the elec

Luda [366]

Answer:

Cost of 1000 kilowatt hour = 6000 cents

Explanation:

Given that

Electricity cost is 6 cents per kilowatt hour.

And we have to found out the cost for one megawatt hour

We know that

1 kilowatt = 1000 watt

1 megawatt = = 1000000 watt

1 megawatt = 1000 kilowatt

1 megawatt hour = 1000 kilowatt hour

Given that cost of 1 kilowatt hour = 6 cents

So the cost of 1000 kilowatt hour = 6 x 1000 cents

Cost of 1000 kilowatt hour = 6000 cents

3 0

2 years ago

A 50.0-g Super Ball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A

Setler [38]

Explanation:

Average acceleration is change in velocity over time.

a = Δv / Δt

a = (22.0 m/s − (-25.0 m/s)) / 0.00350 s

a = 13,400 m/s²

7 0

2 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago