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2 years ago

Why does atmospheric pressure decrease with altitude.

Physics

1 answer:

ella [17]2 years ago

5 0

Answer:

<em>Earth's gravity pulls air as close to the surface as possible. ... As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense than air nearer to sea level. This is what meteorologists and mountaineers mean by "thin air." Thin air exerts less pressure than air at a lower altitude.</em>

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The velocity of an object is the distance it travels per unit time. Suppose the velocity of a gilding bird is measured to be 52.

Elanso [62]

Answer:

$d=7.115s$

Explanation:

What problem says can be written mathematically as:

$v=\frac{d}{t}$

Where:

$v=Velocity\\t=Time\\d=Distance$

The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:

$52\frac{cm}{s} *\frac{1m}{100cm} =0.52\frac{m}{s}$

Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:

$0.52=\frac{3.7}{t}$

Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:

$t=\frac{3.7}{0.52} =7.115384615s\approx7.115s$

5 0

3 years ago

In the picture at left. The launch height is 10m ball B mass 5kg has a launch velocity of 12 m/s and ball A mass 8 kg has a laun

Hitman42 [59]

Answer:

Explanation:

8 0

3 years ago

A bicyclist with a mass of 50 kg is traveling at a rate of 30 m/s. It accelerates to a rate of 50 m/s in 5 seconds. What is the

balandron [24]

Answer:

F=m*(v^2/r)

F=82*(8^2/30)

F=174.9N

Explanation:

brainlest pls

8 0

3 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

3 years ago

The law of universal gravitation (gravity) describes two factors that affect the force of gravity between objects. What two fact

kykrilka [37]

The force depends on the mass of both objects and the distance between them

F = G*m1*m2/r^2

So the force has a linear connection with the mass of both objects and a quadratic connection with the distance between the center of masses

4 0

3 years ago