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Kay [80]
3 years ago
15

It is said that a gas fills all the space available to it. Why then doesn't the atmosphere go off into space?

Physics
1 answer:
Alex777 [14]3 years ago
7 0
Earth has its own atmosphere. That is one reason all the water that has been on Earth has been recycled through the water cycle. It never leaves Earth’s atmosphere.
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HELP PLEASE <br> Find the net force necessary for a 15 kg object to accelerate at 20 m/s/s.
tensa zangetsu [6.8K]

Answer:

<h3>The answer is 300 N</h3>

Explanation:

The force acting on an object given the mass and acceleration we use the formula

<h3>force = mass × acceleration</h3>

We have

force = 15 × 20

We have the final answer as

<h3>300 N</h3>

Hope this helps you

3 0
3 years ago
A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.
evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

where

(A) a = -k × v

 at v= 910 ft/s

     a = - 1.987 × 10⁶ ft/s²

(B)  at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{dv}{v} = -kdt

\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt

ln(200.2)-ln(900) = -kt

t = 6.84 × 10⁻⁴ s

3 0
3 years ago
1. What happens to particle spacing when temperature increase, What we call this process?
enot [183]

Answer:

Particle spacing increases and it's called evaporating

5 0
4 years ago
Science: Look at the image and answer the question correctly.
Flauer [41]
I believe it is D I’m sorry if it’s wrong
4 0
3 years ago
To determine the coefficient of static friction between two materials, an engineer places a small sample of one material on a ho
Fudgin [204]

This question is incomplete, the missing image is uploaded along this answer.

Answer:

the coefficient of friction is 0.32

Explanation:

 Given the data in the question;

we make use of kinematic equation of motion;

ω = ω₀ + ∝t

we substitute

ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )

ω = 3.9612 rad/s

The centripetal force acting on the sample is;

Fc = mrω²

from the image; r = 200 mm = 0.2 m

so we substitute

Fc = m(0.2 m ) ( 3.9612 rad/s )²

Fc = (3.13822 m/s²)m

we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;

f = Fc

μN = Fc

μmg = (3.13822 m/s²)m

μ = (3.13822 m/s²)m / mg

μ = (3.13822 m/s²) / g

acceleration due to gravity g = 9.8 m/s²

so

μ = (3.13822 m/s²) / 9.8 m/s²

μ = 0.32

Therefore, the coefficient of friction is 0.32

5 0
3 years ago
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