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3 years ago

Which layer of the atmosphere has the highest density?

Physics

1 answer:

fomenos3 years ago

4 0

Answer:

The Troposphere is the lowermost portion of Earth’s atmosphere. It is the densest layer of the atmosphere and contains approximately 75 percent of the mass of the atmosphere and almost all the water vapor and aerosol. The troposphere extends from the Earth’s surface up to the tropopause where the stratosphere begins.

Explanation:

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What is the momentum of a 50-kilogram ice skater gliding across the ice at a speed of 5 m/s? (1 point)

serious [3.7K]

Momentum is a term used to quantify the motion of an object has. It is calculated as the the product of the object's mass and the velocity. It is expressed as:

Momentum = m x v
Momentum = 50 kg x 5 m/s
Momentum = 250 kg m/s

Therefore, the correct answer is the last option.

3 0

3 years ago

Read 2 more answers

A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

Arlecino [84]

Answer:

Mathematically:

That will be

= 1500 x 15 x 4500

= 101,250,000

5 0

2 years ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

3 years ago

A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an

ella [17]

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

= 79.68 * 0.0683

= 5.02