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Lostsunrise [7]
3 years ago
15

At the surface of Venus the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is

92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarthgEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude.
a. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
b. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.
c. What is the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km?

Physics
1 answer:
yanalaym [24]3 years ago
8 0

Answer:

a) 86 atm

b) 86 atm

c) 645 m/s

Explanation:

See attachment for calculations on how i arrived at the answer

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A ramp leading to the entrance of a building is inclied upward at an angle of 7 degree. A suitcase is to be pulled up the ramp b
photoshop1234 [79]

Explanation:

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / F

F = 50 / cos 28°

F = 56.6 lbs

8 0
3 years ago
In order to catch a fast-moving softball with your bare hand, you extend your hand forward just before the catch and then let th
hichkok12 [17]
This is a concept of momentum. In equation, momentum is the product of force and distance. When a ball is thrown, its force is constant all throughout unless disturbed by an external force. Therefore, force is the constant of proportionality that relates momentum with distance. When you block a ball from a given distance, you would feel the great force on your hand. In order to reduce the force, you have to follow the direction of the force in order to minimize the impact. By doing this, you gradually decrease the momentum of the ball. 
8 0
3 years ago
How do kinetic and potential energy transfer to one throughout a roller coaster ride?
mojhsa [17]

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

6 0
2 years ago
Please answer this question for me and explain why.
horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}

Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0
3 years ago
The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o
Anuta_ua [19.1K]

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

E_n=\frac{-13.6eV}{n^2}

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

E_{k}=11.1eV-10.2eV=0.9eV

6 0
3 years ago
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