All categories

3 years ago

On which terrestrial planet is the largest volcano in the solar system located? A. Mars B. Earth C. Venus D. Mercury

2 answers:

6 0

The answer is A.Mars

Olympus Mons is the name of the largest volcano in the solar system, and It is located in Mars.

dlinn [17]3 years ago

5 0

It’s not mars it’s earth

You might be interested in

Danny's mother believes her own fear of big dogs may have led to his phobia. This explanation is pointing to ______________.

matrenka [14]

<h2><u>Answer:</u></h2>

Cynophobia

<h3><u>Explanation:</u></h3>

Cynophobia originates from the Greek words that signify "dog" (cyno) and "fear" (phobis). An individual who has cynophobia encounters a dread of mutts that is both unreasonable and tenacious. It's something beyond feeling of scaredness whether a dog is barking or an individual is around dogs.

An individual who has cynophobia encounters a dread of dogs that is both silly and constant. Explicit fears, similar to cynophobia, influence somewhere in the range of 7 to 9 percent of the populace. They're regular enough that they're formally perceived in the Diagnostic and Statistical Manual of Mental Disorders,

6 0

2 years ago

Read 2 more answers

g In 1920, Stern and Gerlach performed an experiment that first demonstrated Group of answer choices energy quantization. space

anyanavicka [17]

Answer:

B. space quantization.

Explanation:

In 1921, Otto Stern developed the idea behind this experiment, while Walther Gerlach performed the actual experiment in 1922. The Ster-Gerlach experiment provides prove to the fact that the spatial orientation of angular momentum is quantized. To demonstrate the experiment, silver atoms were made to travel through a magnetic field path.

Before they hit the screen(usually a glass slide), they were deflected because of their non-zero magnetic moment. There was an expected result for this experiment, but the actual observation on the glass slide was a continuous distribution of the silver atoms that actually hit the glass. This experiment was useful in proving that in all atomic-scale systems, there was a quantization of angular momentum.

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

A car travels 45 km north. It then turns around and travels 30 km south. What is the car's displacement?

natka813 [3]

If the ration supplementary angle is 11:7,find the measure of the larger angle larger angle?

4 0

3 years ago

Biofuels and nuclear power may prove useful as _________.

Papessa [141]

As oil ,coal ,gas ,solar ,wind and water power

5 0

3 years ago