An equilibrium is not changed by a change in pressure.<br> a. True<br> b. False

When does carbon dioxide absorb the most heat energy? uring freezing during deposition during sublimation during condensation?

ololo11 [35]

The answer for the given question above would be the third option. Carbon dioxide absorbs the most heat energy during SUBLIMATION. By definition, sublimation is <span>the transition of a substance from the solid to the gas phase without passing through the intermediate liquid phase. Hope this answers your question.</span>

3 0

3 years ago

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(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

According to Newton's third law, if you push on a wall, the wall ?

Lady bird [3.3K]

The wall will push back, in exactly the opposite direction, and with
exactly the same size force.

That's why the net force on the palm of your hand is zero, and that
in turn is the reason that your hand doesn't accelerate.

If you keep increasing the strength of your push, then eventually you
exceed the force that the wall is capable of delivering. Then the wall
crumbles and falls, your hand accelerates in the direction you're pushing,
and the crowd goes wild !

3 0

3 years ago

Read 2 more answers

An object is placed in front of a convex mirror with a radius of curvature of magnitude 10 cm. The mirror produces an image that

jek_recluse [69]

Answer:

u = - 20 cm

m = $\frac{1}{5}$

Given:

Radius of curvature, R = 10 cm

image distance, v = 4 cm

Solution:

Focal length of the convex mirror, f:

f = $\frac{R}{2} = \frac{10}{2} = 5 cm$

Using Lens' maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Substitute the given values in the above formula:

$\frac{1}{5} = \frac{1}{u} + \frac{1}{4}$

$\frac{1}{u} = \frac{1}{5} - \frac{1}{4}$

u = - 20 cm

where

u = object distance

Now, magnification is the ratio of image distance to the object distance:

magnification, m = $\frac{|v|}{|u|}$

magnification, m = $\frac{|4|}{|-20|}$

m = $\frac{4}{20}$

m = $\frac{1}{5}$

4 0

3 years ago

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

WINSTONCH [101]

Answer:

Explanation:

frequency of sound waves = 688 Hz

wavelength = 344 / 688 = .5 m

The problem is based on interference of sound waves

For the observer , path difference of sound waves reaching his ear

= 3.5 - 3.00

.5 m

= wavelength

Path difference is equal to wavelength so there will be constructive interference and hence louder sound will be heard by the listener than normal sound as sound waves interfere constructively.

6 0

3 years ago

An equilibrium is not changed by a change in pressure. a. True b. False