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3 years ago

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9. A car accelerates from 3.5 m/s to 17 m/s in 4.5 seconds. Find the acceleration of the car in m/s2.

1 answer:

Citrus2011 [14]3 years ago

8 0

The answer is -3 m/s2

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Using knowledge of states of matter,write a message about the importance of science in our society

oee [108]

Answer:

उव्ग्वुव ह्व्झ एउएइहे एइएइएइएएइ सिसुब्स्सी बीस सिस इस्ब एइब

Explanation:

?उग्व्ब्वु विब्सिए इसिग्व विद्बिअब्द सिह्व्व इस्ब्व दिव्ब्स विह्द ऐद्जिइ सुउगव्दी सिइगैगे क्ज्गैइव अजिव्व्ज्व्स कैह्द अजि ह्ज्फ्ज इअह इकुगै ईग इअबे अजिव्ब जैइअब इऐहे ऐइहे ऐइग्गे अत्व्ब ओप्झब रोज दिधिए ऊइफ्ब इसुहद ईउहे सिउउअ दिइब्द स्सिउए ऐइहे सिएय्व एउविये एइव्वे

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2 years ago

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777dan777 [17]

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3 years ago

Two everyday examples of a starting motion

Tasya [4]

Pushing, pulling is the answer

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3 years ago

Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a

Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

$P = \frac{F}{A}$

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

$P = \frac{40\ N}{0.03\ m^2}\\\\$

<u>P = 1333.33 N</u>

4 0

3 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