Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
now, Pressure at depth x
integrating both side
now,
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
Answer:There are two main types of friction, static friction and kinetic friction. Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.
The net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.
To find the answer, we have to know more about the basic forces acting on a body.
<h3>How to find the net force on the box?</h3>
- Let us draw the free body diagram of the given box with the data's given in the question.
- From the diagram, we get,
where, N is the normal reaction, mg is the weight of the box, 
is the net force, f is the kinetic friction.
- We have the expression for kinetic friction as,
- Thus, the net force will be,
Thus, we can conclude that, the net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.
Learn more about the basic forces on a body here:
brainly.com/question/28061293
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Answer:
Option D: 1.5in in front of the target
Explanation:
The object distance is 
.
Because the surface is flat, the radius of curvature is infinity .
The incident index is 
and the transmitted index is 
.
The single interface equation is 
Substituting the quantities given in the problem,
The image distance is then 
Therefore, the coin falls 
in front of the target
