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3 years ago

Why is atomaspheric pressure greater at the surface on Earth

2 answers:

8 0

Because there are miles of air on top of you and air has mass so all the mass of the air on top of you is on your head and the ground. (the air can't penetrate the ground)

Take note: Air is also denser at sea level because it was compressed.

andrew-mc [135]3 years ago

5 0

Hello!

Because as you get closer to the surface of the earth, the more air that is on top of you. At the top of the atmosphere, there is less air, and everything is a vacuum, where you have no weight. When you get close to the earth, the weight of the air builds until it when you're at the very lowest point of the earths surface, all the air in the atmosphere above you is pressing down.

Thank You!

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For each of the problems below, you will need to draw a graph to find the solution.

Gelneren [198K]

Answer:

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

Explanation:

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2 years ago

Big bang theorists believe that the universe is expanding and will eventually contract.

Zina [86]

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3 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

kupik [55]

Answer:

2hrs and some mins

Explanation:

bc 2×36= 17 =)

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3 years ago

A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the

const2013 [10]

Answer:

The equation for the object's displacement is $u(t)=0.583cos11.35t$

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

$k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in$

The angular speed is:

$w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s$

The damping force is:

$F_{D} =cu$

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

$c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in$

The critical damping is equal:

$c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in$

Like cc>c the system is undamped

The equilibrium expression is:

$u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t$

3 0

3 years ago

5. Each of five satellites makes a circular orbit about an object that is much more massive than any of the satellites. The mass

Vikentia [17]

The options of the question are missing. I have attached it.

Answer:

Satellite in option B will have the greatest speed.

Explanation:

From kepplers third law, we know that

V² = GM/R

Thus, v = √(GM/R)

Where;

v is velocity

G is gravitational constant

M is mass

R is radius

Looking at the options, let's start from the first one;

Option A

Here, mass = (1/2)m and radius = R

So, v = √(GM/R) thus v = √(G(m/2)/R) = √(Gm/2R)

Option B

Here, mass = m and radius = (1/2)R

Thus v = √(Gm/(R/2)) = √(2Gm/R)

Option C

Here, mass = m and radius = R

Thus v = √(Gm/R)

Option D

Here, mass = m and radius = 2R

Thus v = √(Gm/(2R))

Now, inspecting all the options, it's clear that option B will have the greatest velocity because it's numerator is the biggest and will in turn lead to higher velocity.

4 0

3 years ago