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3 years ago

Less massive molecules tend to escape from an atmosphere more often than more massive ones because

1 answer:

6 0

We can confirm that less massive molecules tend to escape from an atmosphere more often than more massive ones because they are moving faster.

<h3>How does speed help molecules escape?</h3>

This has to do with the energy of the molecules. Speed comes with its higher kinetic energy. This higher level of energy helps the molecules to escape by giving them enough energy to overpower the force of gravity acting upon them in the atmosphere.

Therefore, we can confirm that less massive molecules tend to escape from an atmosphere more often than more massive ones because they are moving faster.

To learn more about molecules visit:

brainly.com/question/19922822?referrer=searchResults

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Which of the following is an engineering solution

Alona [7]

By "solution" it means a course of action that, once carried out, brings about some desired state of affairs. The use of engineer in this context is as a verb meaning "to arrange or bring about through skillful, artful contrivance."

5 0

4 years ago

Read 2 more answers

A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass

arlik [135]

Answer:

a) $v_{60^{o}} =4.98 m/s$

b) $\theta_{max}=79.34^{o}$

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

$K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}$

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

$K_{lowest}=K_{60^{o}}+U_{60^{0}}$

So now, we can substitute the respective equations for kinetic and potential energy so we get:

$\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}$

we can divide both sides of the equation into the mass of the pendulum so we get:

$\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}$

and we can multiply both sides of the equation by 2 to get:

$v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}$

so we can solve this for $v_{60^{o}}$ . So we get:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

$cos \theta = \frac{y}{4m}$

so we solve for y to get:

$y = 4cos \theta$

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

$h_{60^{o}}=4m-2m$

$h_{60^{o}}=2m$

So now we can substitute the data in the velocity equation we got before:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

$v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}$

$v_{60^{o}}=4.98 m/s$

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

$K_{lowest}+U_{lowest}=K_{max}+U_{max}$

In this case, we know that $U_{lowest}$ will be zero and $K_{max}$ will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

$K_{lowest} =U_{max}$

And now we substitute for the respective kinetic energy and potential energy equations.

$\frac{1}{2}mv_{lowest}^{2}=mgh_{max}$

again, we can divide both sides of the equation into the mass, so we get:

$\frac{1}{2}v_{lowest}^{2}=gh_{max}$

and solve for the height:

$h_{max}=\frac{v_{lowest}^{2}}{2g}$

and substitute:

$h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}$

to get:

$h_{max}=3.26m$

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

$cos \theta_{max}= \frac{0.74}{4}$

$\theta_{max}=cos^{-1}(\frac{0.74}{4})$

so we get:

$\theta_{max}=79.34^{o}$

5 0

3 years ago

A force of attraction would exist between

siniylev [52]

Answer:

A. positively charged object and a negatively charged object.

3 0

4 years ago

Jorge rides his bicycle to school every morning. The school is 4 km away from Jorge's house and it takes him 30 minutes to ride

Liono4ka [1.6K]

D. Speed
"Jorge's Average SPEED is 8km/hr"
hope I helped! also when doing multiple choice try using process of elimination. helps alot :P

5 0

3 years ago

the young’s modulus of aluminum is 69gpa, of nylon is 3gpa, of tungsten is 400gpa, and of copper is 117gpa. if equal-size sample

olga_2 [115]

Answer:

Least to most elongated: tungsten, copper, aluminum, nylon.

Explanation:

Materials with high Young's modulus are difficult to stretch. σ = Yε and ε = ΔL/L so an object with a high Young's modulus (Y) subject to a certain tensile stress (σ) will have a smaller strain than an object with a smaller Young's 's modulus subject to the same tensile stress. If strain (ε) is smaller, then ΔL will also be smaller.

3 0

2 years ago