What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist?


Answer 1




In order to retain atmosphere a planet needs to have gravity. A gravity sufficient enough to create a dense atmosphere around it, so that it can retain heat coming from sun. Mars has shallow atmosphere as its gravity is only 40% of the Earth's gravity. Venus is somewhat similar to Earth but due to green house effect its temperature is very high. Atmosphere has a huge impact on the planets ability to sustain life. Presence of certain kind gases make the atmosphere poisnous for life. The atmosphere should be such that it allows water to remain in liquid form and maintain an optimum temperature suitable for life.

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A gate with a circular cross section is held closed by a lever 1 m long attached to a buoyant cylinder. The cylinder is 25 cm in diameter and weighs 200 N. The gate is attached to a horizontal shaft so it can pivot about its center. The liquid is water. The chain and lever attached to the gate have negligible weight. Find the length of the chain such that the gate is just on the verge of opening when the water depth above the gate hinge is 10 m.


The length of the chain such that the gate is just on the verge of opening  is mathematically given as


What is the length of the chain?

Generally, the equation for the   is mathematically given as

F'=\gamma hA


9810*10* (\pi)/(4)1^2

Fh= 77048 N


ycp-y=((\pi r^4)/(4))/(10*(\pi D^4)/(4))


In conclusion, resultant force

x = F'' - W

x = 9810* 10*( \pi/4 )*0.25^2 *(10-l)-200

x = 4615.5-481.5 l


77048* 0.00625 - 1 *(4615.5-481.5 l) = 0


Read more about Lenght





length of lever = 1 m

diameter of cylinder = 25 cm

weight of cylinder = 200 N

hydrostatic force

F_H=\gamma hA

      = 9810* 10* (\pi)/(4)1^2

      = 77048 N


y_(cp)-\bar{y}= \frac{I}{\bar{y}A}

                    =  ((\pi r^4)/(4))/(10*(\pi D^4)/(4))

                    =  0.00625 m

Finding the resultant force

F = F_(buoyancy) - W

F = \gamma_(water)A(10-l)-W

F = 9810* 10* (\pi)/(4)0.25^2* (10-l)-200

F = 4615.5-481.5 l

taking moment about hinge

F_(H)* 0.00625 - 1 * F = 0

77048* 0.00625 - 1 *(4615.5-481.5 l) = 0

l = 8.58 m

A mysterious object with a surface area of 0.015 m2, volume of 0.000125 m3, density of 100 kg/m3, specific heat of 100 J/(kgK), thermal conductivity of 2 W/(mK), with an unknown initial temperature was placed in a fluid with a density of 50 kg/m3, specific heat of 70 J/(kgK), thermal conductivity of 0.1 W/(mK), at a temperature of 400K. The heat transfer coefficient is given to be 10 W/(m2K). After 10 seconds, the temperature of the object is measured to be 380K. Determine the object's initial temperature.



The object's initial temperature is 333.6 K


We first assume that the liquid can only transfer heat to the object through convective heat transfer method.

Let T₀ = the initial temperature of the object

T = temperature of the object at anytime.

The rate of heat transfer from the liquid to the object is given as

Q = -hA (T∞ - T)

T∞ = temperature of the fluid = 400 K

A = Surface area of the object in contact with the liquid = 0.015 m²

h = Convective heat transfer coefficient is given to be = 10 W/(m²K)

The rate of heat gained by the object is given by

mC (d/dt)(T∞ - T)

m = mass of the object = ρV

ρ = density of the object = 100 kg/m³

V = volume of the object = 0.000125 m³

m = ρV = 100 × 0.000125 = 0.0125 kg

C = specific heat capacity of the object = 100 J/(kgK)

The rate of heat loss by the liquid = rate of heat gain by the object

-hA (T∞ - T) = mC (d/dt)(T∞ - T)

(d/dt)(T∞ - T) = - (dT/dt) ( Since T∞ is a constant)

- mC (dT/dt) = -hA (T∞ - T)

(dT/dt) = (hA/mC) (T∞ - T)

Let s = (hA/mC)

(dT/dt) = -s (T - T∞)

dT/(T - T∞) = -sdt

Integrating the left hand side from T₀ (the initial temperature of the object) to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -st

(T - T∞)/(T₀ - T∞) = e⁻ˢᵗ

(T - T∞) = (T₀ - T∞)e⁻ˢᵗ

s = (hA/mC) = (10 × 0.015)/(0.0125×100) = 0.12

T = 380 K at t = 10 s

T₀ = ?

T∞ = 400 K

st = 0.12 × 10 = 1.2

(380 - 400) = (T₀ - 400) e⁻¹•²

(-20/0.3012) = (T₀ - 400)

(T₀ - 400) = - 66.4

T₀ = 400 - 66.4 = 333.6 K

Hope this Helps!!!

As frequency increases in an electromagnetic wave, its photon energy decreases. A. True B. False









In a physics laboratory experiment, a coil with 170 turns enclosing an area of 10.9 cm2 is rotated during the time interval 3.50×10−2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.60×10−5 T. What is the total magnitude of the magnetic flux (initial) through the coil before it is rotated?


N= 170 turns\nA=10.9cm^2\nt=3.5*10^(-2)s \n\B = 5.6*10^(-5)T

The Magnetic flow \Phi_(initial) is given by the formula,


Replacing the values

\Phi_(Initial) =(5.6*10^(-5))(10.9)((10^(-4)m^2)/(1cm^2)) sin90\°

\Phi_(Initial) =6.104*10^(-7) Wb

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.


To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

m_1v_1+m_2v_2 = (m_1+m_2)v_f


m_(1,2)= Mass of each object

v_(1,2) = Initial Velocity of each object

v_f= Final velocity

Replacing we have that,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

1850*13.8+3100*0 = (1850+3100)v_f

v_f = 5.1575m/s

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

v_f^2-v_i^2 = 2ax

Since there is no initial speed, then

v_f^2 = 2ax

5.1575^2 = 2a (1.91)

a = 6.9633m/s^2

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

F_f = F_a \n\mu N = m*a \n\mu = (ma)/(N)\n\mu = (ma)/(mg)\n\mu = (a)/(g)\n\mu = (6.9633)/(9.8)\n\mu = 0.7105

Therefore the Kinetic friction coefficient is 0.7105

A spectroscope breaks light up into its colors, allowing scientists to analyze light from the solar system and universe. By studying the spectral line patterns, widths, strengths and positions, scientists can determine the speed, position, and _____ of celestial bodies.A) age
B) origin
C) rotation
D) temperature


It is D - temperature

Final answer:

A spectroscope analyses light to determine various parameters of celestial bodies. The missing parameter in this context is the 'temperature' of the celestial body (option D). The spectral lines, based on their pattern and strengths helps in determining this.


A spectroscope decomposes or breaks white light into its spectrum of colors, allowing scientists to study them and understand various aspects of celestial bodies. When scientists analyze the spectral line patterns, widths, strengths, and positions, they can discern essential parameters. These parameters include the speed and position of the celestial body, and more importantly, the correct answer to your question, its temperature (option D). This is because every element when heated, absorbs or emits light at characteristic wavelengths, that give us the 'spectral lines'. By studying these we can determine the temperature of the celestial body.

Learn more about Spectroscope here: