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3 years ago

We dont got the force so i got no clue what to do

Physics

1 answer:

konstantin123 [22]3 years ago

3 0

Answer:

work = 1275.3 J

Explanation:

work = (force)(distance)cosø ------- force = ma

=(mass*acceleration)(distance)cosø

=(20*9.81)(6.5)cos0

=1275.3J

nite that the angle of cosine is the difference between the angle of force and the distance. in this case, the force and the distance are in the same direction. :)

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In your egg drop you want to decrease.

Ede4ka [16]

Answer:

Explanation:

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3 years ago

Read 2 more answers

(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

Which description correctly summarizes how an electric motor causes an axle to turn?

alexira [117]

Answer:

B on edge

Explanation:

the piles of the magnetic field generated around the armature are attracted to the opposite poles of the permanent magnent. As the opposite poles align, the commutator reverses the current direction so like poles are aligned and the armature continues to spin

6 0

3 years ago

Q1) An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours.To cover the same distance in 1(2/3) hours, it must

Pepsi [2]

Answer:

D) 720 kmph

Explanation:

First, let's find what distance this aeroplane covered. Distance (d) is the product of speed and time - here, we have a speed of 240 kmph and a time of 5 hours, gives us

$d=240\cdot5=1200$ km

Using that same fact, we can set up a new equation to solve for speed (s) when we have a distance of 1200 km and a time of 1 2/3 hours. For the sake of cleanliness, I'm gonna rewrite 1 2/3 as the improper fraction 5/3:

$1200=(5/3)s$

Multiplying both sides of the equation by 3/5:

$s=(3/5)(1200)=720$ kmph

So our answer is D.

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2 years ago

The astronaut then measures the abundance of silicon on the new planet, obtaining the following results: Isotope Abundance. (%)M

tekilochka [14]

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3 years ago