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3 years ago

Of the planets with atmospheres, which is the warmest? a. Venusb. Earthc. Marsd. Jupiter

1 answer:

7 0

The Atmosphere in Jupiter is full of gases that move at high speeds in giant eddies. Its atmosphere consists mostly of gases such as hydrogen that generate a temperature fluctuation of around 128K.

On Earth, due to the protection of the Ozone Layer and the presence of Nitrogen and Oxygen, the temperature fluctuates by an average of 300K.

In the case of Mars, its atmosphere is thin, mostly composed of Carbon Dioxide and Diatomic Nitrogen, which allow a temperature oscillation of 210K.

In contrast, the atmosphere of Venus is thick and is composed of carbon dioxide that does not allow the sun's rays to escape, generating an extreme 'greenhouse effect' with temperatures ranging from 737K,

Correct Answer is A.

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A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec

seropon [69]

Answer:

$\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )$

Explanation:

The vector that point from point P1 to point P2 its found simply by taking the vector at which point P2 its located and subtracting the vector at which point P1 its located:

$\vec{C} = \vec{B} - \vec{A}$

So:

$\vec{C} = ( \ 1.60 \ , \ 3.66 \ ) - ( \ 3.74 \ , \ 1.64 \ )$

$\vec{C} = ( \ 1.60 \ - \ 3.74 \ , \ 3.66 \ - \ 1.64 \ )$

$\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )$

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2 years ago

A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w

Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0

2 years ago

Read 2 more answers

What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction

fomenos

Complete Question

The complete question is shown on the uploaded image

Answer:

The nuclei experience a force that will move it to the right of the conductor rod while the electrons experience a force that will move it to the left side of the conductor rod.

Explanation:

The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as

F = qE => E = F/q

where F is the force, q is the charge and E is the electric field

we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.

we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge

Now from the question we are being told that the external electric field is the direction of the positive x axis

Hence this field would drive the positive charge i.e the nuclei to the right.

In order to further explain let consider this

Generally the electric field is always radially outward originating from a positive point charge and radially in toward a negative point charge.

what this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.

According to Coulomb's law which states that unlike term attract while like terms repel, it means that the electron would move to the left of the conductor rod while the nuclei would move to the right side of the conductor rod.