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?

Answers

Answer 1
Answer:

Answer:

Explained

Explanation:

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.


Related Questions

For the system shown below, what is the critical angle (angle at which the system just begins to move)? Assume that the coefficient of friction between all flat surfaces is 0.0500 and that the pulley is frictionless. The mass of m1 is 76.00 kg and the mass of m2 is 194.00 kg. Express your answer in radians.
An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places
The fundamental force that is responsible for beta decay is the: A) weak force. B) strong force. C) dark energy force. D) gravitational force E) electromagnetic force.
Exposure to what type of radiant energy is sensed by human skin as warmth? x-rays ultraviolet infrared gamma rays
In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after are given. The collision is (Show your work-no work shown = ZERO POINTS) 1.8 m/s 0.2 m/s 0.6 m/s 1.4 m/s 4 kg 6 kg 4 kg 6 kg Before After A) perfectly elastic. B) partially inelastic. C) completely inelastic. D) characterized by an increase in kinetic energy E) not possible because momentum is not conserved.

How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 J
B) 5 J
C) 50 J
D) 1 J
E) 10 J

Answers

Answer:

option C

Explanation:

given,                            

Force on the object = 10 N

distance of push = 5 m

Work done = ?              

we know,              

work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

Final answer:

The work done on an object when a force of 10 N pushes it 5 m is 50 Joules, calculated by multiplying the force and the displacement. So, the correct option is C.

Explanation:

The question is asking about work, which in physics is the result of a force causing a displacement. The formula for work is defined as the product of the force (in Newtons) and the displacement (in meters) the force causes. If a force of 10 N pushes an object a distance of 5 m, the work done is calculated by multiplying the force and the displacement (10 N * 5 m), yielding 50 Joules of work.

Therefore, the correct answer is 50 J (C).

Learn more about Work here:

brainly.com/question/31965083

#SPJ6

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) v=5.6725\,m.s^(-1)

b) h= 1.6420\,m

Explanation:

Given:

  • mass of the body, M=20\,kg
  • mass of the tyre,m=10\,kg
  • length of hanging of tyre, l=3.5m
  • distance run by the body, d=10m
  • acceleration of the body, a=3.62m.s^(-2)

(a)

Using the equation of motion :

v^2=u^2+2a.d..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

u=0m.s^(-1)

putting the values in eq. (1)

v^2=0^2+2* 3.62 * 10

v=8.5088\,m.s^(-1)

Now, the momentum of the body just before the jump onto the tyre will be:

p=M.v

p=20* 8.5088

p=170.1764\,kg.m.s^(-1)

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

(M+m)* v'=p

(20+10)* v'=170.1764

v'=5.6725\,m.s^(-1)

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

(1)/(2) (M+m).v'^2=(M+m).g.h

(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h

h\approx 1.6420\,m

above the normal hanging position.

When a tennis ball is spun around in a circle on a string and the string breaks the tennis ballwill be pulled in a curved path away from the center because of Centrifugal force
True or false

Answers

Answer:

False

whenever the string breaks, the ball will follow the straight line tangential path

Explanation:

No, the ball will not follow a curved path after the string breaks. Since, the the direction of velocity is tangential to each point of the circular motion. Therefore, it changes at every point. This produces an acceleration in the circle called centripetal acceleration. There is also a tangential component of acceleration acting on the ball during this motion.

So, whenever the string breaks, the ball will follow the straight line tangential path. Hence, the given statement is false.

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy the equation PV=C, where C is a constant. Suppose that at a certain instant the volume is 600cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 20 kPa/min. At what rate is the volume decreasing at this instant?

Answers

Answer: The volume is decreasing at a rate of 80 cm3/min

Explanation: Please see the attachments below

Answer: 80 cm³/min

Explanation:

Just solved it

The speed of light in a transparent plastic is 2.5x108 m/s. If a ray of light in air(with n-1) strikes this plastic at an angle of incidence of 31.3 degrees, find the angle of the transmitted ray. Select one: a. 31.30 degrees b. 26.08 degrees c, 37.56 degrees d. 38.57 degrees e. 0.67 degrees

Answers

Answer:

Angle of transmitted ray is 29.14^(o)

Explanation:

According to snell's law we have

n_(1)sin(\theta _(i))=n_(2)sin(\theta r)

Since the incident medium is air thus we have n_(1)=1

By definition of refractive index we have

n=(c)/(v)

c = speed of light in vacuum

v = speed of light in medium

Applying values we get

n_(2)=(3* 10^(8))/(2.5* 10^(8))=1.2

Thus using the calculated values in Snell's law we obtain

sin(\theta _(r))=(sin(31.3)/(1.2)\n\n\therefore sin(\theta _(r))=0.4329\n\n\theta_(r)=sin^(-1)(0.4329)\n\n\theta _(r)=29.14

Answer:

Angle made by the transmitted ray = 25.65°

Explanation:

Speed of light in plastic = v = 2.5 × 10⁸ m/s  

refractive index of plastic (n₂) / refractive index of air (n₁)

= speed of light in air c / speed of light in plastic v.

⇒ n₂ = (3× 10⁸) / (2.5 × 10⁸) = 1.2

Angle of incidence = 31.3° = i  

n₁ sin i = n₂ sin r

⇒  sin r =  (1)(0.5195) / 1.2 = 0.4329

⇒ Angle made by the transmitted ray = r = sin⁻¹ (0.4329) =25.65°

             

A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answers

Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3